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In the previous exercises, we learned how to evaluate integrals involving sums and differences of functions, apply the method of substitution, and integrate rational functions using partial fractions. But what should we do when the integrand is the product of two functions?
\displaystyle \int x e^x\,dx,\qquad \int x\sin x\,dx,\qquad \int (\log x)^2\,dx
Such integrals are generally evaluated using the Method of Integration by Parts. This method is based on differentiating one function while integrating the other, thereby reducing the given integral to a simpler form.
In words, the formula can be stated as:
The integral of the product of two functions = (First function) × (Integral of the second function) − Integral of [(Derivative of the first function) × (Integral of the second function)].
In this exercise, we will learn the formula for Integration by Parts, understand how to choose the first and second functions using the ILATE Rule, study its important exceptions, and solve different types of questions step by step.
Key Concepts
1. Integration by Parts Formula
Integration by Parts is used to evaluate integrals in which the integrand is the product of two functions.
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
Here, u represents the first function and v represents the second function. The first function is differentiated, whereas the second function is integrated.
2. Choosing the First Function (ILATE Rule)
The success of this method depends on choosing u correctly. A useful guideline is the ILATE Rule, which gives the preferred order for selecting the first function.
- I – Inverse Trigonometric Functions
- L – Logarithmic Functions
- A – Algebraic Functions
- T – Trigonometric Functions
- E – Exponential Functions
Generally, we choose u as the first function from the type of function appearing earlier in the above order. The remaining function is taken as v, the second function.
3. Special Type
Some integrals involving exponential functions can be solved directly using a special result instead of applying Integration by Parts repeatedly.
\displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^x f(x)+C
Recognising this pattern can save considerable time in calculations.
4. Important Points to Remember
Keep the following points in mind while applying the method of Integration by Parts.
- Choosing the first and second functions correctly is important. A wrong choice may lead to a more difficult integral. For example, while evaluating \displaystyle \int x\cos x\,dx, choosing x as the first function makes the solution much simpler.
- Not every product of two functions can be integrated by Parts. For example, \displaystyle \int \sqrt{x}\sin x\,dx cannot be evaluated using this method because there is no function whose derivative is \sqrt{x}\sin x.
- Some integrals require Integration by Parts to be applied more than once. For example, \displaystyle \int e^x\sin x\,dx and \displaystyle \int e^x\cos x\,dx are evaluated by applying the method twice.
- If the integrand consists of only one function, we may treat it as the product of that function and the constant function 1 (unity). In such cases, the given function is taken as the first function and 1 as the second function. For example, \displaystyle \int \log x\,dx,\ \int \sin^{-1}x\,dx,\ \int (\log x)^2\,dx and \displaystyle \int (\sin^{-1}x)^2\,dx are evaluated by writing the integrand as \displaystyle \log x\cdot1,\ \sin^{-1}x\cdot1,\ (\log x)^2\cdot1 and \displaystyle (\sin^{-1}x)^2\cdot1, respectively.
- While applying Integration by Parts, do not add the constant of integration during intermediate steps. Add the constant C only once after obtaining the final answer.
5. Strategy for Solving Integration by Parts Questions
- Check whether the integrand is the product of two functions.
- Choose the first function using the ILATE Rule, whenever applicable.
- The remaining function becomes the second function.
- Apply the Integration by Parts formula.
- Simplify the resulting integral and repeat the process if required.
- Add the constant of integration C only once at the end.
Following these steps provides a systematic approach for solving most Integration by Parts questions in this exercise.
Tip: In Integration by Parts, the choice of the first function is often the key to the solution. Use the ILATE Rule as a guideline, simplify the resulting integral whenever possible, and add the constant of integration C only once at the end.
Quick Reference: Integration by Parts
The following table summarises the most important concepts, formulas and strategies for quick revision of Integration by Parts. Go through it once before solving the exercise and again before your exam for a quick recap.
| Quick Reference: Integration by Parts |
|---|
| Integration by Parts Formula |
| \displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx |
| Easy Recall |
| \displaystyle \int \mathrm{I}\cdot\mathrm{II}\,dx=\mathrm{I}\int \mathrm{II}\,dx-\int\left(\frac{d(\mathrm{I})}{dx}\int \mathrm{II}\,dx\right)dx |
| ILATE Rule |
| I – Inverse Trigonometric Functions |
| L – Logarithmic Functions |
| A – Algebraic Functions |
| T – Trigonometric Functions |
| E – Exponential Functions |
| Generally, we choose u as the first function from the type of function appearing earlier in the above order. The remaining function is taken as v, the second function. |
| Special Type |
| \displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C |
| Recognising this pattern helps avoid repeated application of Integration by Parts. |
Note: Some integrals, such as \displaystyle \int e^x\sin x\,dx and \displaystyle \int e^x\cos x\,dx, require Integration by Parts to be applied twice. These are also exceptions to the ILATE Rule because either function may be chosen as the first function, and both approaches eventually lead to the correct answer.
Let us now solve all the NCERT questions step by step in this exercise of Integrals 7.6.
Question 1: Integrals 7.6
1. Evaluate \displaystyle \int x\sin x\,dx.
Solution
Let, \displaystyle I=\int x\sin x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose x as the first function and \sin x as the second function.
So, considering \displaystyle I=\int \underbrace{x}_{\mathrm{I}}\;\underbrace{\sin x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=x\int\sin x\,dx-\int\left(\frac{d}{dx}(x)\int\sin x\,dx\right)dx
Since \displaystyle \int\sin x\,dx=-\cos x and \displaystyle \frac{d}{dx}(x)=1, we get
\displaystyle I=x(-\cos x)-\int\left(1\cdot(-\cos x)\right)dx
Simplifying, we get
\displaystyle I=-x\cos x+\int\cos x\,dx
Using \displaystyle \int\cos x\,dx=\sin x, we obtain
\displaystyle I=-x\cos x+\sin x+C
Hence,
\boxed{\displaystyle \int x\sin x\,dx=-x\cos x+\sin x+C}
Question 2: Integrals Ex. 7.6
2. Evaluate \displaystyle \int x\sin3x\,dx.
Solution
Let, \displaystyle I=\int x\sin3x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose x as the first function and \sin3x as the second function.
So, considering \displaystyle I=\int \underbrace{x}_{\mathrm{I}}\;\underbrace{\sin3x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=x\int\sin3x\,dx-\int\left(\frac{d}{dx}(x)\int\sin3x\,dx\right)dx
Since \displaystyle \int\sin3x\,dx=-\frac{\cos3x}{3} and \displaystyle \frac{d}{dx}(x)=1, we get
\displaystyle I=x\left(-\frac{\cos3x}{3}\right)-\int\left(1\cdot-\frac{\cos3x}{3}\right)dx
Simplifying, we get
\displaystyle I=-\frac{x\cos3x}{3}+\frac{1}{3}\int\cos3x\,dx
Using \displaystyle \int\cos3x\,dx=\frac{\sin3x}{3}, we obtain
\displaystyle I=-\frac{x\cos3x}{3}+\frac{\sin3x}{9}+C
Hence,
\boxed{\displaystyle \int x\sin3x\,dx=-\frac{x\cos3x}{3}+\frac{\sin3x}{9}+C}
You may already be following Maths Better for NCERT Solutions for the topics like
- Matrices
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Likewise Integrals 7.6 for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills. Now, let’s proceed to the next question.
Question 3: Integration by Parts
3. Evaluate \displaystyle \int x^2e^x\,dx.
Solution
Let, \displaystyle I=\int x^2e^x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose x^2 as the first function and e^x as the second function.
So, considering \displaystyle I=\int\underbrace{x^2}_{\mathrm{I}}\;\underbrace{e^x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=x^2\int e^x\,dx-\int\left(\frac{d}{dx}(x^2)\int e^x\,dx\right)dx
Since \displaystyle \int e^x\,dx=e^x and \displaystyle \frac{d}{dx}(x^2)=2x, we get
\displaystyle I=x^2e^x-\int2xe^x\,dx
The remaining integral is again the product of two functions. Therefore, we apply Integration by Parts once more.
Let, \displaystyle I_1=\int2xe^x\,dx
Using the ILATE Rule, we choose 2x as the first function and e^x as the second function.
So, considering \displaystyle I_1=\int\underbrace{2x}_{\mathrm{I}}\;\underbrace{e^x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I_1=2x\int e^x\,dx-\int\left(\frac{d}{dx}(2x)\int e^x\,dx\right)dx
Since \displaystyle \int e^x\,dx=e^x and \displaystyle \frac{d}{dx}(2x)=2, we get
\displaystyle I_1=2xe^x-\int2e^x\,dx
Using \displaystyle \int e^x\,dx=e^x, we obtain
\displaystyle I_1=2xe^x-2e^x
Substituting this value of I_1 in the expression for I, we get
\displaystyle I=x^2e^x-(2xe^x-2e^x)+C
Simplifying, we get
\displaystyle I=e^x(x^2-2x+2)+C
Hence,
\boxed{\displaystyle \int x^2e^x\,dx=e^x(x^2-2x+2)+C}
Question 4: Integrals 7.6
4. Evaluate \displaystyle \int x\log x\,dx.
Solution
Let, \displaystyle I=\int x\log x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \log x as the first function and x as the second function.
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
So, considering \displaystyle I=\int\underbrace{\log x}_{\mathrm{I}}\;\underbrace{x}_{\mathrm{II}}\,dx
We get, \displaystyle I=\log x\int x\,dx-\int\left(\frac{d}{dx}(\log x)\int x\,dx\right)dx
Since \displaystyle \int x\,dx=\frac{x^2}{2} and \displaystyle \frac{d}{dx}(\log x)=\frac{1}{x}, we get
\displaystyle I=\frac{x^2\log x}{2}-\int\left(\frac{1}{x}\cdot\frac{x^2}{2}\right)dx
Simplifying, we get
\displaystyle I=\frac{x^2\log x}{2}-\frac{1}{2}\int x\,dx
Using \displaystyle \int x\,dx=\frac{x^2}{2}, we obtain
\displaystyle I=\frac{x^2\log x}{2}-\frac{x^2}{4}+C
Hence,
\boxed{\displaystyle \int x\log x\,dx=\frac{x^2}{2}\log x-\frac{x^2}{4}+C}
Question 5: Method of Integration by Parts
5. Evaluate \displaystyle \int x\log(2x)\,dx.
Solution
Let, \displaystyle I=\int x\log(2x)\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \log(2x) as the first function because logarithmic functions are preferred over algebraic functions. Hence, x becomes the second function.
So, considering \displaystyle I=\int\underbrace{\log(2x)}_{\mathrm{I}}\;\underbrace{x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\log(2x)\int x\,dx-\int\left(\frac{d}{dx}(\log(2x))\int x\,dx\right)dx
Since \displaystyle \int x\,dx=\frac{x^2}{2} and \displaystyle \frac{d}{dx}(\log(2x))=\frac{1}{2x}\cdot2=\frac{1}{x}, we get
\displaystyle I=\frac{x^2\log(2x)}{2}-\int\left(\frac{1}{x}\cdot\frac{x^2}{2}\right)dx
Simplifying, we get
\displaystyle I=\frac{x^2\log(2x)}{2}-\frac{1}{2}\int x\,dx
Using \displaystyle \int x\,dx=\frac{x^2}{2}, we obtain
\displaystyle I=\frac{x^2\log(2x)}{2}-\frac{x^2}{4}+C
Hence,
\boxed{\displaystyle \int x\log(2x)\,dx=\frac{x^2}{2}\log(2x)-\frac{x^2}{4}+C}
Tip: From Q. 4 and Q. 5, notice that \displaystyle \frac{d}{dx}(\log x)=\frac{1}{x} and \displaystyle \frac{d}{dx}(\log(2x))=\frac{1}{2x}\cdot2=\frac{1}{x}. In general, by the Chain Rule,
\displaystyle \frac{d}{dx}(\log(ax))=\frac{1}{ax}\cdot a=\frac{1}{x},\qquad a\ne0
where a is a non-zero constant.
Question 6: ILATE Rule: Integration by Parts
6. Evaluate \displaystyle \int x^2\log x\,dx.
Solution
Let, \displaystyle I=\int x^2\log x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \log x as the first function because logarithmic functions are preferred over algebraic functions. Hence, x^2 becomes the second function.
So, considering \displaystyle I=\int\underbrace{\log x}_{\mathrm{I}}\;\underbrace{x^2}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\log x\int x^2\,dx-\int\left(\frac{d}{dx}(\log x)\int x^2\,dx\right)dx
Since \displaystyle \int x^2\,dx=\frac{x^3}{3} and \displaystyle \frac{d}{dx}(\log x)=\frac{1}{x}, we get
\displaystyle I=\frac{x^3\log x}{3}-\int\left(\frac{1}{x}\cdot\frac{x^3}{3}\right)dx
Simplifying, we get
\displaystyle I=\frac{x^3\log x}{3}-\frac{1}{3}\int x^2\,dx
Using \displaystyle \int x^2\,dx=\frac{x^3}{3}, we obtain
\displaystyle I=\frac{x^3\log x}{3}-\frac{x^3}{9}+C
Hence,
\boxed{\displaystyle \int x^2\log x\,dx=\frac{x^3}{3}\log x-\frac{x^3}{9}+C}
Question 7: NCERT Exercise Integrals 7.6
7. Evaluate \displaystyle \int x\sin^{-1}x\,dx.
Solution
Let, \displaystyle I=\int x\sin^{-1}x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \sin^{-1}x as the first function because inverse trigonometric functions are preferred over algebraic functions. Hence, x becomes the second function.
So, considering \displaystyle I=\int\underbrace{\sin^{-1}x}_{\mathrm{I}}\;\underbrace{x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\sin^{-1}x\int x\,dx-\int\left(\frac{d}{dx}(\sin^{-1}x)\int x\,dx\right)dx
Since \displaystyle \int x\,dx=\frac{x^2}{2} and \displaystyle \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}, we get
\displaystyle I=\frac{x^2\sin^{-1}x}{2}-\frac12\int\frac{x^2}{\sqrt{1-x^2}}\,dx
or \displaystyle I=\frac{x^2\sin^{-1}x}{2}-I_1\qquad\cdots(1)
where, \displaystyle I_1=\int\frac{x^2}{\sqrt{1-x^2}}\,dx
Let’s write the numerator as \displaystyle x^2=1-(1-x^2), so that we can split the integral into two standard forms.
\displaystyle \therefore I_1=\int\frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx
Splitting the integral, we get
\displaystyle I_1=\int\frac{dx}{\sqrt{1-x^2}}-\int\frac{1-x^2}{\sqrt{1-x^2}}\,dx
Since \displaystyle \frac{1-x^2}{\sqrt{1-x^2}}=\sqrt{1-x^2}, we get
\displaystyle I_1=\int\frac{dx}{\sqrt{1-x^2}}-\int\sqrt{1-x^2}\,dx
Using \displaystyle \int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x and \displaystyle \int\sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x, we obtain
\displaystyle I_1=\sin^{-1}x-\left(\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x\right)
Simplifying, we get
\displaystyle I_1=\frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}
Substituting the value of I_1 in equation (1), we get
\displaystyle I=\frac{x^2\sin^{-1}x}{2}-\frac12\left(\frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}\right)+C
Simplifying, we get
\displaystyle I=\frac{(2x^2-1)\sin^{-1}x}{4}+\frac{x}{4}\sqrt{1-x^2}+C
Hence,
\boxed{\displaystyle I=\frac14(2x^2-1)\sin^{-1}x+\frac{x\sqrt{1-x^2}}{4}+C}
Note: While evaluating the above integral, we used the standard result
\displaystyle \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C
This is one of the important special integrals that will be discussed in detail in Exercise 7.7. The same result can also be obtained by substituting x=a\sin\theta.
Question 8: Integrals 7.6
8. Evaluate \displaystyle \int x\tan^{-1}x\,dx.
Solution
Let, \displaystyle I=\int x\tan^{-1}x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \tan^{-1}x as the first function because inverse trigonometric functions are preferred over algebraic functions. Hence, x becomes the second function.
So, considering \displaystyle I=\int\underbrace{\tan^{-1}x}_{\mathrm{I}}\;\underbrace{x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\tan^{-1}x\int x\,dx-\int\left(\frac{d}{dx}(\tan^{-1}x)\int x\,dx\right)dx
Since \displaystyle \int x\,dx=\frac{x^2}{2} and \displaystyle \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}, we get
\displaystyle I=\frac{x^2\tan^{-1}x}{2}-\frac12\int\frac{x^2}{1+x^2}\,dx
or \displaystyle I=\frac{x^2\tan^{-1}x}{2}-\frac12I_1\qquad\cdots(1)
where, \displaystyle I_1=\int\frac{x^2}{1+x^2}\,dx
Let’s write the numerator as \displaystyle x^2=(1+x^2)-1, so that we can split the integral into two simpler integrals.
\displaystyle \therefore I_1=\int\frac{(1+x^2)-1}{1+x^2}\,dx
Splitting the integral, we get
\displaystyle I_1=\int\frac{1+x^2}{1+x^2}\,dx-\int\frac{dx}{1+x^2}
Since \displaystyle \frac{1+x^2}{1+x^2}=1, we get
\displaystyle I_1=\int dx-\int\frac{dx}{1+x^2}
Using \displaystyle \int dx=x and \displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x, we obtain
\displaystyle I_1=x-\tan^{-1}x
Substituting the value of I_1 in equation (1), we get
\displaystyle I=\frac{x^2\tan^{-1}x}{2}-\frac12\left(x-\tan^{-1}x\right)+C
or \displaystyle I=\frac{x^2}{2}\tan^{-1}x-\frac{x}{2}+\frac12\tan^{-1}x+C
Hence,
\boxed{\displaystyle I=\frac{x^2}{2}\tan^{-1}x-\frac{x}{2}+\frac12\tan^{-1}x+C}
Congratulations on completing NCERT Class 12 Maths Part 1! We have covered all the chapters and exercises with detailed explanations and step-by-step solutions. The journey continues with Part 2, where I’ll keep providing easy-to-follow solutions and concept-based explanations. Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, to help you learn and revise more effectively.
Now, let’s proceed to the next question of Integrals 7.6.
Question 9: Integration by Parts
9. Evaluate \displaystyle \int x\cos^{-1}x\,dx.
Solution
Let, \displaystyle I=\int x\cos^{-1}x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \cos^{-1}x as the first function because inverse trigonometric functions are preferred over algebraic functions. Hence, x becomes the second function.
So, considering \displaystyle I=\int\underbrace{\cos^{-1}x}_{\mathrm{I}}\;\underbrace{x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\cos^{-1}x\int x\,dx-\int\left(\frac{d}{dx}(\cos^{-1}x)\int x\,dx\right)dx
Since \displaystyle \int x\,dx=\frac{x^2}{2} and \displaystyle \frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}}, we get
\displaystyle I=\frac{x^2\cos^{-1}x}{2}+\frac12\int\frac{x^2}{\sqrt{1-x^2}}\,dx
or \displaystyle I=\frac{x^2\cos^{-1}x}{2}+\frac12I_1\qquad\cdots(1)
where, \displaystyle I_1=\int\frac{x^2}{\sqrt{1-x^2}}\,dx
Let’s write the numerator as \displaystyle x^2=1-(1-x^2), so that we can split the integral into two standard forms.
\displaystyle \therefore I_1=\int\frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx
Splitting the integral, we get
\displaystyle I_1=\int\frac{dx}{\sqrt{1-x^2}}-\int\frac{1-x^2}{\sqrt{1-x^2}}\,dx
Since \displaystyle \frac{1-x^2}{\sqrt{1-x^2}}=\sqrt{1-x^2}, we get
\displaystyle I_1=\int\frac{dx}{\sqrt{1-x^2}}-\int\sqrt{1-x^2}\,dx
Using \displaystyle \int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x and \displaystyle \int\sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x, we obtain
\displaystyle I_1=\sin^{-1}x-\left(\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x\right)
Simplifying, we get
\displaystyle I_1=\frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}
Substituting the value of I_1 in equation (1), we get
\displaystyle I=\frac{x^2\cos^{-1}x}{2}+\frac12\left(\frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}\right)+C
Simplifying, we get
\displaystyle I=\frac{x^2}{2}\cos^{-1}x+\frac14\sin^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C
Using the identity \displaystyle \sin^{-1}x=\frac{\pi}{2}-\cos^{-1}x, we get
\displaystyle \frac14\sin^{-1}x=\frac{\pi}{8}-\frac14\cos^{-1}x
Substituting this in the above expression, we get
\displaystyle I=\frac{x^2}{2}\cos^{-1}x+\frac{\pi}{8}-\frac14\cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C
Simplifying, we get
\displaystyle I=\frac14(2x^2-1)\cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C'
where \displaystyle C'=C+\frac{\pi}{8}. Hence,
\boxed{\displaystyle I=\frac14(2x^2-1)\cos^{-1}x-\frac{x\sqrt{1-x^2}}{4}+C'}
Question 10: Integrals 7.6
10. Evaluate \displaystyle \int(\sin^{-1}x)^2\,dx.
Solution
Let, \displaystyle I=\int(\sin^{-1}x)^2\,dx
Since the integrand contains only one function, we write it as the product of (\sin^{-1}x)^2 and 1. We then use the Method of Integration by Parts.
Using the ILATE Rule, we choose (\sin^{-1}x)^2 as the first function and 1 as the second function.
So, considering \displaystyle I=\int\underbrace{(\sin^{-1}x)^2}_{\mathrm{I}}\;\underbrace{1}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=(\sin^{-1}x)^2\int dx-\int\left(\frac{d}{dx}(\sin^{-1}x)^2\int dx\right)dx
Since \displaystyle \int dx=x and \displaystyle \frac{d}{dx}(\sin^{-1}x)^2=\frac{2\sin^{-1}x}{\sqrt{1-x^2}}, we get
\displaystyle I=x(\sin^{-1}x)^2-2\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx
or \displaystyle I=x(\sin^{-1}x)^2-2I_1\qquad\cdots(1)
where, \displaystyle I_1=\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx
Using the substitution t=\sin^{-1}x, we get
\displaystyle x=\sin t,\qquad dx=\cos t\,dt,\qquad \sqrt{1-x^2}=\cos t
Therefore,
\displaystyle I_1=\int t\sin t\,dt
Again using Integration by Parts,
\displaystyle I_1=t(-\cos t)-\int(-\cos t)\,dt
\displaystyle I_1=-t\cos t+\sin t+C
Substituting \displaystyle t=\sin^{-1}x,\;\cos t=\sqrt{1-x^2},\;\sin t=x, we get
\displaystyle I_1=-\sin^{-1}x\sqrt{1-x^2}+x
Substituting the value of I_1 in equation (1), we get
\displaystyle I=x(\sin^{-1}x)^2-2\left(-\sin^{-1}x\sqrt{1-x^2}+x\right)+C
Simplifying, we get
\displaystyle I=x(\sin^{-1}x)^2+2\sqrt{1-x^2}\sin^{-1}x-2x+C
Hence,
\boxed{\displaystyle I=x(\sin^{-1}x)^2+2\sqrt{1-x^2}\sin^{-1}x-2x+C}
Tip: Integration by Parts and Substitution are often used together. After applying Integration by Parts, the remaining integral may not always be evaluated directly. In such cases, look for a suitable substitution that simplifies the integrand. In Q. 10 and Q. 11, substituting x=sin t or \sin\theta converts the remaining integral into a much simpler form.
Question 11: Integrals 7.6
11. Evaluate \displaystyle \int\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx.
Solution
Let, \displaystyle I=\int\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \cos^{-1}x as the first function because inverse trigonometric functions are preferred over algebraic functions. Hence, \dfrac{x}{\sqrt{1-x^2}} becomes the second function.
So, considering \displaystyle I=\int\underbrace{\cos^{-1}x}_{\mathrm{I}}\;\underbrace{\frac{x}{\sqrt{1-x^2}}}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\cos^{-1}x\int\frac{x}{\sqrt{1-x^2}}\,dx-\int\left(\frac{d}{dx}(\cos^{-1}x)\int\frac{x}{\sqrt{1-x^2}}\,dx\right)dx
or \displaystyle I=\cos^{-1}x\,I_1+\int\frac{I_1}{\sqrt{1-x^2}}\,dx\qquad\cdots(1)
where, \displaystyle I_1=\int\frac{x}{\sqrt{1-x^2}}\,dx
Using the substitution t=1-x^2, we get
\displaystyle dt=-2x\,dx\qquad\Rightarrow\qquad x\,dx=-\frac12\,dt
Therefore,
\displaystyle I_1=-\frac12\int\frac{dt}{\sqrt{t}}
Using \displaystyle \int t^{-1/2}\,dt=2\sqrt{t}, we get
\displaystyle I_1=-\sqrt{t}
Substituting t=1-x^2, we obtain
\displaystyle I_1=-\sqrt{1-x^2}
Substituting the value of I_1 in equation (1), we get
\displaystyle I=-\sqrt{1-x^2}\cos^{-1}x+\int\frac{-\sqrt{1-x^2}}{\sqrt{1-x^2}}\,dx
Simplifying, we get
\displaystyle I=-\sqrt{1-x^2}\cos^{-1}x-\int dx
Using \displaystyle \int dx=x, we obtain
\displaystyle I=-\sqrt{1-x^2}\cos^{-1}x-x+C
Hence,
\boxed{\displaystyle I=-\sqrt{1-x^2}\cos^{-1}x-x+C}
Alternative Method
The integral \displaystyle I_1=\int\frac{x}{\sqrt{1-x^2}}\,dx can also be evaluated by using the substitution x=\sin\theta.
Using x=\sin\theta, we get
\displaystyle dx=\cos\theta\,d\theta,\qquad \sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\cos\theta
Therefore,
\displaystyle I_1=\int\frac{\sin\theta}{\cos\theta}\cdot\cos\theta\,d\theta=\int\sin\theta\,d\theta
Integrating, we get
\displaystyle I_1=-\cos\theta+C
Substituting \displaystyle \theta=\sin^{-1}x, we obtain
\displaystyle I_1=-\sqrt{1-x^2}+C
Question 12: Method of Integration by Parts
12. Evaluate \displaystyle \int x\sec^2x\,dx.
Solution
Let, \displaystyle I=\int x\sec^2x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose x as the first function and \sec^2x as the second function.
So, considering \displaystyle I=\int\underbrace{x}_{\mathrm{I}}\;\underbrace{\sec^2x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=x\int\sec^2x\,dx-\int\left(\frac{d}{dx}(x)\int\sec^2x\,dx\right)dx
Since \displaystyle \int\sec^2x\,dx=\tan x and \displaystyle \frac{d}{dx}(x)=1, we get
\displaystyle I=x\tan x-\int\tan x\,dx
Using \displaystyle \int\tan x\,dx=\log|\sec x|, we obtain
\displaystyle I=x\tan x-\log|\sec x|+C
Hence,
\displaystyle I=x\tan x-\log|\sec x|+C
or we can write it as
\boxed{\displaystyle I=x\tan x+\log|\cos x|+C}
Question 13: Integrals 7.6
13. Evaluate \displaystyle \int\tan^{-1}x\,dx.
Solution
Let, \displaystyle I=\int\tan^{-1}x\,dx
Since the integrand contains only one function, we write it as the product of \tan^{-1}x and 1. We then use the Method of Integration by Parts.
Using the ILATE Rule, we choose \tan^{-1}x as the first function because inverse trigonometric functions are preferred over constant functions. Hence, 1 becomes the second function.
So, considering \displaystyle I=\int\underbrace{\tan^{-1}x}_{\mathrm{I}}\;\underbrace{1}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\tan^{-1}x\int dx-\int\left(\frac{d}{dx}(\tan^{-1}x)\int dx\right)dx
Since \displaystyle \int dx=x and \displaystyle \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}, we get
\displaystyle I=x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx
Using the substitution t=1+x^2, we get
\displaystyle dt=2x\,dx\qquad\Rightarrow\qquad x\,dx=\frac12\,dt
Therefore,
\displaystyle \int\frac{x}{1+x^2}\,dx=\frac12\int\frac{dt}{t}=\frac12\log|t|+C
Substituting t=1+x^2, we get
\displaystyle \int\frac{x}{1+x^2}\,dx=\frac12\log(1+x^2)
Hence,
\displaystyle I=x\tan^{-1}x-\frac12\log(1+x^2)+C
Therefore,
\boxed{\displaystyle I=x\tan^{-1}x-\frac12\log(1+x^2)+C}
Tip: If the integrand contains only one function, write it as the product of that function and the constant function 1. Then apply Integration by Parts by taking the given function as the first function and 1 as the second function. This technique is used in Q. 10 and Q. 13.
Question 14: Integrals Exercise 7.6
14. Evaluate \displaystyle \int x(\log x)^2\,dx.
Solution
Let, \displaystyle I=\int x(\log x)^2\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose (\log x)^2 as the first function because logarithmic functions are preferred over algebraic functions. Hence, x becomes the second function.
So, considering \displaystyle I=\int\underbrace{(\log x)^2}_{\mathrm{I}}\;\underbrace{x}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=(\log x)^2\int x\,dx-\int\left(\frac{d}{dx}(\log x)^2\int x\,dx\right)dx
Since \displaystyle \int x\,dx=\frac{x^2}{2} and \displaystyle \frac{d}{dx}(\log x)^2=\frac{2\log x}{x}, we get
\displaystyle I=\frac{x^2(\log x)^2}{2}-\int x\log x\,dx
or \displaystyle I=\frac{x^2(\log x)^2}{2}-I_1\qquad\cdots(1)
where, \displaystyle I_1=\int x\log x\,dx
As we have done in Q. 4, (to show the same steps here) we get,
\displaystyle I_1=\frac{x^2}{2}\log x-\frac{x^2}{4}+C
Substituting the value of I_1 in equation (1), we get
\displaystyle I=\frac{x^2(\log x)^2}{2}-\left(\frac{x^2}{2}\log x-\frac{x^2}{4}\right)+C
Simplifying, we get
\displaystyle I=\frac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C
Hence,
\boxed{\displaystyle I=\frac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C}
Question 15: Indefinite Integrals 7.6
15. Evaluate \displaystyle \int (x^2+1)\log x\,dx.
Solution
Let, \displaystyle I=\int(x^2+1)\log x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \log x as the first function because logarithmic functions are preferred over algebraic functions. Hence, x^2+1 becomes the second function.
So, considering \displaystyle I=\int\underbrace{\log x}_{\mathrm{I}}\;\underbrace{(x^2+1)}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\log x\int(x^2+1)\,dx-\int\left(\frac{d}{dx}(\log x)\int(x^2+1)\,dx\right)dx
Since \displaystyle \int(x^2+1)\,dx=\frac{x^3}{3}+x and \displaystyle \frac{d}{dx}(\log x)=\frac1x, we get
\displaystyle I=\left(\frac{x^3}{3}+x\right)\log x-\int\left(\frac{x^3/3+x}{x}\right)dx\qquad\cdots(1)
or \displaystyle I=\left(\frac{x^3}{3}+x\right)\log x-I_1\qquad\cdots(1)
where, \displaystyle I_1=\int\frac{x^3/3+x}{x}\,dx
Simplifying the integrand, we get
\displaystyle I_1=\int\left(\frac{x^2}{3}+1\right)dx
Using the linearity property of integration, we get
\displaystyle I_1=\frac13\int x^2\,dx+\int dx
Using \displaystyle \int x^2\,dx=\frac{x^3}{3} and \displaystyle \int dx=x, we obtain
\displaystyle I_1=\frac{x^3}{9}+x
Substituting the value of I_1 in equation (1), we get
\displaystyle I=\left(\frac{x^3}{3}+x\right)\log x-\left(\frac{x^3}{9}+x\right)+C
Simplifying, we get
\displaystyle I=\left(\frac{x^3}{3}+x\right)\log x-\frac{x^3}{9}-x+C
Hence,
\boxed{\displaystyle I=\left(\frac{x^3}{3}+x\right)\log x-\frac{x^3}{9}-x+C}
Tip: In questions involving e^x, first check whether the given expression can be written in the form e^x\left(f(x)+f'(x)\right). If this pattern is recognised, the integral can be evaluated directly using the result
\displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C
Otherwise, apply the Method of Integration by Parts. The next few questions illustrate both approaches.
Question 16: Special Type of Integrals 7.6
16. Evaluate \displaystyle \int e^x(\sin x+\cos x)\,dx.
Solution
Let, \displaystyle I=\int e^x(\sin x+\cos x)\,dx
Here, the given integrand is of the form e^x(f(x)+f'(x)), where f(x)=\sin x. Although this integral can be evaluated directly using the special result, let us first derive that result by applying the Method of Integration by Parts.
Using the linearity property of integration, we get
\displaystyle I=\int e^x\sin x\,dx+\int e^x\cos x\,dx\qquad\cdots(1)
Applying Integration by Parts to the first integral only, we choose \sin x as the first function and e^x as the second function. So, considering
\displaystyle I=\int\underbrace{\sin x}_{\mathrm{I}}\;\underbrace{e^x}_{\mathrm{II}}\,dx+\int e^x\cos x\,dx
Using \displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx in the first integral, we get
\displaystyle \int e^x\sin x\,dx=e^x\sin x-\int e^x\cos x\,dx
Substituting this in equation (1), we get
\displaystyle I=e^x\sin x-\int e^x\cos x\,dx+\int e^x\cos x\,dx
The two integrals cancel each other. Therefore,
\displaystyle I=e^x\sin x+C
Hence,
\boxed{\displaystyle I=e^x\sin x+C}
This proves the special result
\displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C
Important: The special result \displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C can be used only when the given expression is exactly of the form e^x\left(f(x)+f'(x)\right). Even a change in the coefficient changes the result. For example, e^x(\sin x+2\cos x) is not exactly of this form because 2\cos x\neq\dfrac{d}{dx}(\sin x). However, we can first write
\displaystyle e^x(\sin x+2\cos x)=e^x(\sin x+\cos x)+e^x\cos x
Now the special result can be applied to the first integral, while the remaining integral \displaystyle \int e^x\cos x\,dx is evaluated using the Method of Integration by Parts.
Question 17: Special Type of Integrals 7.6
17. Evaluate \displaystyle \int\frac{xe^x}{(1+x)^2}\,dx.
Solution
Let, \displaystyle I=\int\frac{xe^x}{(1+x)^2}\,dx
The given integrand is not directly of the form e^x\left(f(x)+f'(x)\right). Therefore, we first simplify the algebraic expression so that the special result can be applied.
Let’s write the numerator as \displaystyle x=(1+x)-1. Then,
\displaystyle I=\int\frac{\big((1+x)-1\big)e^x}{(1+x)^2}\,dx
Splitting the numerator, we get
\displaystyle I=\int e^x\left(\frac1{1+x}-\frac1{(1+x)^2}\right)\,dx\qquad\cdots(1)
Now, let \displaystyle f(x)=\frac1{1+x}. Then,
\displaystyle f'(x)=-\frac1{(1+x)^2}
Hence,
\displaystyle \frac1{1+x}-\frac1{(1+x)^2}=f(x)+f'(x)
Thus, the given integrand is reduced to the form e^x\left(f(x)+f'(x)\right)
Therefore, using the special result
\displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C
and from equation (1), we get
\displaystyle I=\frac{e^x}{1+x}+C
Hence,
\boxed{\displaystyle \int\frac{xe^x}{(1+x)^2}\,dx=\frac{e^x}{1+x}+C}
Question 18: Integrals 7.6
18. Evaluate \displaystyle \int e^x\left(\frac{1+\sin x}{1+\cos x}\right)\,dx.
Solution
Let, \displaystyle I=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)\,dx
The given integrand is not directly of the form e^x\left(f(x)+f'(x)\right). Therefore, we first simplify the trigonometric expression so that the special result can be applied.
Splitting the numerator, we get
Using the identities \displaystyle 1+\cos x=2\cos^2\frac x2 and \displaystyle \sin x=2\sin\frac x2\cos\frac x2, we get
\displaystyle I=\int e^x\left(\frac{1}{2\cos^2\frac x2}+\frac{2\sin\frac x2\cos\frac x2}{2\cos^2\frac x2}\right)\,dx
Simplifying, we get
\displaystyle I=\int e^x\left(\frac12\sec^2\frac x2+\tan\frac x2\right)\,dx\qquad\cdots(1)
Now, let \displaystyle f(x)=\tan\frac x2. Then,
\displaystyle f'(x)=\frac12\sec^2\frac x2
Hence,
\displaystyle \tan\frac x2+\frac12\sec^2\frac x2=f(x)+f'(x)
Thus, the given integrand is reduced to the form e^x\left(f(x)+f'(x)\right).
Therefore, using the special result
\displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C
and from equation (1), we get
\displaystyle I=e^x\tan\frac x2+C
Hence,
\boxed{\displaystyle \int e^x\left(\frac{1+\sin x}{1+\cos x}\right)\,dx=e^x\tan\frac x2+C}
Question 19: Integrals Exercise 7.6
19. Evaluate \displaystyle \int e^x\left(\frac1x-\frac1{x^2}\right)\,dx.
Solution
Let, \displaystyle I=\int e^x\left(\frac1x-\frac1{x^2}\right)\,dx
The given integrand is already of the form e^x\left(f(x)+f'(x)\right).
Let, \displaystyle f(x)=\frac1x. Then,
\displaystyle f'(x)=-\frac1{x^2}
Hence,
\displaystyle \frac1x-\frac1{x^2}=f(x)+f'(x)
Therefore, using the special result
\displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C
we get
\displaystyle I=\frac{e^x}{x}+C
Hence,
\boxed{\displaystyle \int e^x\left(\frac1x-\frac1{x^2}\right)\,dx=\frac{e^x}{x}+C}
Question 20: Special Integrals 7.6
20. Evaluate \displaystyle \int\frac{(x-3)e^x}{(x-1)^3}\,dx.
Solution
Let, \displaystyle I=\int\frac{(x-3)e^x}{(x-1)^3}\,dx
The given integrand is not directly of the form e^x\left(f(x)+f'(x)\right). Therefore, we first simplify the algebraic expression so that the special result can be applied.
Let’s write the numerator as \displaystyle x-3=(x-1)-2. Then,
\displaystyle I=\int\frac{\big((x-1)-2\big)e^x}{(x-1)^3}\,dx
Splitting the numerator, we get
\displaystyle I=\int e^x\left(\frac1{(x-1)^2}-\frac2{(x-1)^3}\right)\,dx\qquad\cdots(1)
Now, let \displaystyle f(x)=\frac1{(x-1)^2}. Then,
\displaystyle f'(x)=-\frac2{(x-1)^3}
Hence,
\displaystyle \frac1{(x-1)^2}-\frac2{(x-1)^3}=f(x)+f'(x)
Thus, the given integrand is reduced to the form e^x\left(f(x)+f'(x)\right).
Therefore, using the special result
\displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C
and from equation (1), we get
\displaystyle I=\frac{e^x}{(x-1)^2}+C
Hence,
\boxed{\displaystyle \int\frac{(x-3)e^x}{(x-1)^3}\,dx=\frac{e^x}{(x-1)^2}+C}
Tip: The expression inside the brackets may not always be directly of the form f(x)+f'(x). Before applying the special result, first try to reduce it to the required form by using suitable algebraic manipulation or trigonometric identities. In Q. 17 and Q. 20, we simplify the algebraic expression, whereas in Q. 18, we use trigonometric identities to obtain the required form. Once the expression is reduced to f(x)+f'(x), the special result can be applied directly.
Question 21: Exponential Integrals 7.6
21. Evaluate \displaystyle \int e^{2x}\sin x\,dx.
Solution
Let, \displaystyle I=\int e^{2x}\sin x\,dx
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose \sin x as the first function and e^{2x} as the second function.
So, considering \displaystyle I=\int\underbrace{\sin x}_{\mathrm{I}}\;\underbrace{e^{2x}}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
Since \displaystyle \int e^{2x}\,dx=\frac12e^{2x} and \displaystyle \frac{d}{dx}(\sin x)=\cos x, we get
\displaystyle I=\frac12e^{2x}\sin x-\frac12\int e^{2x}\cos x\,dx
or \displaystyle I=\frac12e^{2x}\sin x-\frac12I_1\qquad\cdots(1)
where, \displaystyle I_1=\int e^{2x}\cos x\,dx
Again applying the Method of Integration by Parts, we choose \cos x as the first function and e^{2x} as the second function.
So, considering \displaystyle I_1=\int\underbrace{\cos x}_{\mathrm{I}}\;\underbrace{e^{2x}}_{\mathrm{II}}\,dx
We get \displaystyle I_1=\frac12e^{2x}\cos x+\frac12\int e^{2x}\sin x\,dx
Since \displaystyle \int e^{2x}\sin x\,dx=I, we get
\displaystyle I_1=\frac12e^{2x}\cos x+\frac12I
Substituting the value of I_1 in equation (1), we get
\displaystyle I=\frac12e^{2x}\sin x-\frac12\left(\frac12e^{2x}\cos x+\frac12I\right)
Simplifying, we get
\displaystyle I=\frac12e^{2x}\sin x-\frac14e^{2x}\cos x-\frac14I
Taking \displaystyle \frac14I to the left side, we get
\displaystyle \frac54I=\frac12e^{2x}\sin x-\frac14e^{2x}\cos x
Multiplying both sides by \displaystyle \frac45, we obtain
\displaystyle I=\frac{2}{5}e^{2x}\sin x-\frac15e^{2x}\cos x+C
Hence,
\boxed{\displaystyle \int e^{2x}\sin x\,dx=\frac{e^{2x}}5(2\sin x-\cos x)+C}
Question 22: Substitution Method – Integrals 7.6
22. Evaluate \displaystyle \int\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\,dx.
Solution
Let, \displaystyle I=\int\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\,dx
Using the substitution x=\tan\theta, we get
\displaystyle dx=\sec^2\theta\,d\theta
Also,
\displaystyle \frac{2x}{1+x^2}=\frac{2\tan\theta}{1+\tan^2\theta}=2\sin\theta\cos\theta=\sin2\theta
Therefore,
\displaystyle I=\int2\theta\,\sec^2\theta\,d\theta
Since the integrand is the product of two functions, we use the Method of Integration by Parts.
Using the ILATE Rule, we choose 2\theta as the first function and \sec^2\theta as the second function.
So, considering \displaystyle I=\int\underbrace{2\theta}_{\mathrm{I}}\;\underbrace{\sec^2\theta}_{\mathrm{II}}\,d\theta
Applying the Integration by Parts formula,
\displaystyle \int uv\,d\theta=u\int v\,d\theta-\int\left(\frac{du}{d\theta}\int v\,d\theta\right)d\theta
We get, \displaystyle I=2\theta\int\sec^2\theta\,d\theta-\int\left(\frac{d}{d\theta}(2\theta)\int\sec^2\theta\,d\theta\right)d\theta
Since \displaystyle \int\sec^2\theta\,d\theta=\tan\theta and \displaystyle \frac{d}{d\theta}(2\theta)=2, we get
\displaystyle I=2\theta\tan\theta-\int2\tan\theta\,d\theta
or \displaystyle I=2\theta\tan\theta-2\int\tan\theta\,d\theta
Using \displaystyle \int\tan\theta\,d\theta=\log|\sec\theta|, we obtain
\displaystyle I=2\theta\tan\theta-2\log|\sec\theta|+C
Now, substituting \displaystyle \theta=\tan^{-1}x and \displaystyle \tan\theta=x, we get
\displaystyle \sec\theta=\sqrt{1+\tan^2\theta}=\sqrt{1+x^2}
Therefore,
\displaystyle I=2x\tan^{-1}x-2\log\sqrt{1+x^2}+C
Using the log property \displaystyle n\log m=\log(m^n), we get
\displaystyle I=2x\tan^{-1}x-\log(1+x^2)+C
Hence,
\boxed{\displaystyle I=2x\tan^{-1}x-\log(1+x^2)+C}
Now, let’s solve the two important MCQ’s of Integrals 7.6.
Question 23: Integrals 7.6 – MCQ
23. Choose the correct answer.
\displaystyle \int x^2e^{x^3}\,dx equals
- (A) \displaystyle \frac13e^{x^3}+C
- (B) \displaystyle \frac13e^{x^2}+C
- (C) \displaystyle \frac12e^{x^3}+C
- (D) \displaystyle \frac12e^{x^2}+C
Solution
Let, \displaystyle I=\int x^2e^{x^3}\,dx
Using the substitution x^3=t, we get
\displaystyle dt=3x^2\,dx\qquad\Rightarrow\qquad x^2\,dx=\frac{dt}{3}
Substituting these values, we get
\displaystyle I=\frac13\int e^t\,dt
Using \displaystyle \int e^t\,dt=e^t+C, we obtain
\displaystyle I=\frac13e^t+C
Substituting t=x^3, we get
\boxed{\displaystyle I=\frac13e^{x^3}+C}
✅️ Hence, the correct answer is (A).
Question 24: Integrals 7.6 – MCQ
24. Choose the correct answer.
\displaystyle \int e^x\sec x(1+\tan x)\,dx equals
- (A) \displaystyle e^x\cos x+C
- (B) \displaystyle e^x\sec x+C
- (C) \displaystyle e^x\sin x+C
- (D) \displaystyle e^x\tan x+C
Solution
Let, \displaystyle I=\int e^x\sec x(1+\tan x)\,dx
The given integrand is already of the form e^x\left(f(x)+f'(x)\right).
Let, \displaystyle f(x)=\sec x. Then,
\displaystyle f'(x)=\sec x\tan x
Hence,
\displaystyle \sec x+\sec x\tan x=\sec x(1+\tan x)=f(x)+f'(x)
Therefore, using the special result
\displaystyle \int e^x\left(f(x)+f'(x)\right)\,dx=e^xf(x)+C
we get
\boxed{\displaystyle I=e^x\sec x+C}
✅️ Hence, the correct answer is (B).
Common Mistakes to Avoid
- Choosing the wrong first function: Before applying Integration by Parts, select the first function carefully using the ILATE Rule. A wrong choice may lead to a more complicated integral.
- Applying Integration by Parts to every product: Not every product of two functions can be evaluated using this method. For example, \displaystyle \int\sqrt{x}\sin x\,dx cannot be evaluated by Integration by Parts alone.
- Ignoring a simpler substitution: Before applying Integration by Parts, always check whether a suitable substitution can simplify the integral first, as in questions involving inverse trigonometric functions.
- Not recognising the special form: Expressions of the form \displaystyle e^x\left(f(x)+f'(x)\right) can be integrated directly. Missing this pattern often leads to unnecessary calculations.
- Applying the special result directly without simplification: Sometimes the required form is obtained only after using algebraic manipulation or trigonometric identities. Always simplify the expression first whenever required.
- Making sign mistakes while applying Integration by Parts: Pay special attention to the negative sign in the formula \displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx. A sign error changes the final answer completely.
- Stopping too early: Some integrals require applying Integration by Parts more than once. Continue the process until the integral can be evaluated or the original integral reappears.
- Not solving when the original integral reappears: In some questions, the original integral appears again after applying Integration by Parts twice. Bring all such terms to one side and solve for the required integral.
- Making errors while substituting back: After solving the integral in the substituted variable, always convert the answer completely back to the original variable before writing the final result.
- Forgetting the constant of integration: Every indefinite integral must end with the arbitrary constant C. Omitting it makes the answer incomplete.
- Not checking the final answer: Differentiate your result whenever possible. If differentiation gives back the original integrand, the integration is correct.
Continue Learning
After completing Exercise 7.6 of Integrals, you should now be familiar with the Method of Integration by Parts and know how to apply it in a variety of situations. You have also learned how to choose the correct first function using the ILATE Rule, simplify expressions before applying Integration by Parts, recognise the special form e^x\left(f(x)+f'(x)\right), and combine substitution with Integration by Parts whenever required.
To strengthen your understanding further, make sure that you revise:
- The Integration by Parts formula and how to apply it correctly
- Choosing the correct first function using the ILATE Rule
- Recognising when a constant function (1) should be taken as the second function
- Applying Integration by Parts more than once whenever required
- Solving questions in which the original integral reappears after applying Integration by Parts twice
- Recognising the special form \displaystyle e^x\left(f(x)+f'(x)\right) and reducing expressions to this form whenever required
- Using suitable algebraic manipulation or trigonometric identities before applying the special result
- Identifying when a suitable substitution can simplify the integral before applying Integration by Parts
- Using standard trigonometric identities and logarithmic properties to simplify the final answer whenever required
- Checking the final answer by differentiating it whenever possible
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Integration by Parts becomes much easier with practice. As you solve more questions, first identify the first and second functions using the ILATE Rule, and check whether a suitable substitution or algebraic simplification can make the integration easier. Also, look for special patterns such as \displaystyle e^x\left(f(x)+f'(x)\right), where the integral can be evaluated directly. With regular practice, you’ll quickly recognize common patterns and solve even lengthy integration problems with greater speed and confidence.
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