Integrals: Chapter 7 Links
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Integrals 7.2 introduces some of the most important methods of integration. In many cases, an integral cannot be evaluated directly using the standard formulas learned in Exercise 7.1. To simplify such integrals, we often transform them into a more convenient form using suitable substitutions.
\displaystyle \int f(g(x))\,g'(x)\,dx=\int f(t)\,dt,\qquad \text{where } t=g(x)
The method of substitution is based on the reverse application of the chain rule of differentiation. It is particularly useful when the integrand contains a function along with its derivative, such as f(x)f'(x) or \frac{f'(x)}{f(x)}.
In this exercise, we will learn how to evaluate integrals using suitable substitutions and apply this technique to algebraic, exponential, logarithmic and trigonometric functions. Mastering these basic substitution methods will make many integration problems much easier to solve.
Key Concepts
1. Integration by Substitution
Many integrals become easier when we replace a complicated expression by a new variable. This technique is called integration by substitution and is based on the reverse application of the chain rule of differentiation.
\displaystyle \int f(g(x))\,g'(x)\,dx=\int f(t)\,dt,\qquad \text{where }t=g(x)
The aim is to transform the given integral into a simpler form that can be evaluated using standard integration formulas.
2. Recognising Suitable Substitutions
A suitable substitution is often suggested when the integrand contains a function together with its derivative.
Common patterns:
- f(x)f'(x)
- \displaystyle \frac{f'(x)}{f(x)}
In such cases, choosing t=f(x) usually simplifies the integral significantly.
3. Powers of a Function
A useful extension of the substitution method occurs when the integrand contains a power of a function together with its derivative.
\displaystyle \int [f(x)]^n f'(x)\,dx
In such cases, put t=f(x). The integral then becomes
\displaystyle \int t^n\,dt=\frac{t^{n+1}}{n+1}+C,\qquad n\neq -1
Similarly, integrals of the form
\displaystyle \int \frac{f'(x)}{[f(x)]^n}\,dx
can also be evaluated by the substitution t=f(x).
4. Logarithmic Integrals
Integrals of the form \displaystyle \frac{f'(x)}{f(x)} lead to logarithmic functions and occur frequently in this exercise.
\displaystyle \int \frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C
This result is obtained directly by using the substitution t=f(x).
5. Strategy for Solving Integrals by Substitution
- Look for a function whose derivative is also present in the integrand.
- Choose a suitable substitution such as t=f(x).
- Rewrite the integral completely in terms of the new variable.
- Apply the appropriate integration formula.
- Substitute back in terms of x and add the constant of integration C.
Integration By Substitution
The following patterns appear frequently in this exercise. Whenever you recognize one of them, use the indicated substitution to simplify the integral.
| Pattern | Suggested Substitution |
|---|---|
| \displaystyle \int f(x)\,f'(x)\,dx | Put t=f(x), so dt=f'(x)\,dx gives \int tdt |
| \displaystyle \int \frac{f'(x)}{f(x)}\,dx | Put t=f(x), giving \int \frac{dt}{t} |
| \displaystyle \int [f(x)]^n f'(x)\,dx | Put t=f(x), so that dt=f'(x)\,dx |
| \displaystyle \int \frac{f'(x)}{[f(x)]^n}\,dx | Put t=f(x), giving \int t^{-n}\,dt |
| For example, \displaystyle \int \tan x\,dx | Write \tan x=\frac{\sin x}{\cos x} and use substitution. |
| e.g. \displaystyle \int \cot x\,dx | Write \cot x=\frac{\cos x}{\sin x} and use substitution. |
Important Integration Formulas
The following trigonometric integrals are used frequently in this exercise and should be remembered.
| Integral | Result |
|---|---|
| \displaystyle \int \sin x\,dx | -\cos x+C |
| \displaystyle \int \cos x\,dx | \sin x+C |
| \displaystyle \int \tan x\,dx | \log|\sec x|+C=-\log|\cos x|+C |
| \displaystyle \int \cot x\,dx | \log|\sin x|+C |
| \displaystyle \int \sec x\,dx | \log|\sec x+\tan x|+C |
| \displaystyle \int \cosec x\,dx | \log|\cosec x-\cot x|+C |
| \displaystyle \int \sec^2 x\,dx | \tan x+C |
| \displaystyle \int \cosec^2 x\,dx | -\cot x+C |
| \displaystyle \int \sec x\tan x\,dx | \sec x+C |
| \displaystyle \int \cosec x\cot x\,dx | -\cosec x+C |
Tip: While evaluating an integral, first look for an expression whose derivative (or a constant multiple of it) is also present in the integrand. Such questions can often be solved by a suitable substitution. If the required substitution is not immediately visible, try simplifying or rewriting the expression first. Finally, remember to add the constant of integration C.
Let us now solve all the NCERT questions step by step in this exercise of Integrals 7.2.
Question 1: Integrals 7.2
1. Integrate \displaystyle \frac{2x}{1+x^2}
Solution
This integral is of the form \displaystyle \int \frac{f'(x)}{f(x)}\,dx.
Here, f(x)=1+x^2 and f'(x)=2x. Since the numerator is the derivative of the denominator, we substitute the function f(x) by letting t=1+x^2.
Let
t=1+x^2
Then
dt=2x\,dx
Substituting, we get
\displaystyle \int \frac{2x}{1+x^2}\,dx=\int \frac{dt}{t}
Using the standard result \displaystyle \int \frac{dt}{t}=\log|t|+C, we obtain
\displaystyle \int \frac{dt}{t}=\log|t|+C
Replacing t by 1+x^2, we get
\displaystyle \int \frac{2x}{1+x^2}\,dx=\log|1+x^2|+C
Since 1+x^2>0 for all real x, we may also write
\displaystyle \int \frac{2x}{1+x^2}\,dx=\log(1+x^2)+C
Hence,
\boxed{\displaystyle \int \frac{2x}{1+x^2}\,dx=\log(1+x^2)+C}
Question 2: Integrals 7.2
2. Integrate \displaystyle \frac{(\log x)^2}{x}
Solution
This integral is of the form \displaystyle \int f(x)\,f'(x)\,dx.
Here, f(x)=\log x and f'(x)=\frac{1}{x}. Since the integrand contains a function together with its derivative, we substitute the function f(x) by letting t=\log x.
Let
t=\log x
Then
dt=\frac{1}{x}\,dx
Substituting, we get
\displaystyle \int \frac{(\log x)^2}{x}\,dx=\int t^2\,dt
Using the power rule of integration,
\displaystyle \int t^2\,dt=\frac{t^3}{3}+C
Replacing t by \log x, we get
\displaystyle \int \frac{(\log x)^2}{x}\,dx=\frac{(\log x)^3}{3}+C
Hence,
\boxed{\displaystyle \int \frac{(\log x)^2}{x}\,dx=\frac{(\log x)^3}{3}+C}
You may already be following Maths Better for NCERT Solutions for the Chapters in Part I namely
- Matrices
- Determinants
- Relations and Functions
- Inverse Trigonometric Functions
- Continuity and Differentiability
- Application of Derivatives
Likewise this Integrals 7.2 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 Substitution
3. Integrate \displaystyle \frac{1}{x+x\log x}
Solution
First, simplify the denominator:
\displaystyle \frac{1}{x+x\log x}=\frac{1}{x(1+\log x)}
Therefore,
\displaystyle \int \frac{1}{x+x\log x}\,dx=\int \frac{1}{x(1+\log x)}\,dx
This integral is of the form \displaystyle \int \frac{f'(x)}{f(x)}\,dx.
Here, f(x)=1+\log x and f'(x)=\frac{1}{x}. Since the numerator is the derivative of the denominator, we substitute the function f(x) by letting t=1+\log x.
Let
t=1+\log x
Then
dt=\frac{1}{x}\,dx
Substituting, we get
\displaystyle \int \frac{1}{x(1+\log x)}\,dx=\int \frac{dt}{t}
Using the standard result \displaystyle \int \frac{dt}{t}=\log|t|+C, we obtain
\displaystyle \int \frac{dt}{t}=\log|t|+C
Replacing t by 1+\log x, we get
\displaystyle \int \frac{1}{x+x\log x}\,dx=\log|1+\log x|+C
Hence,
\boxed{\displaystyle \int \frac{1}{x+x\log x}\,dx=\log|1+\log x|+C}
Question 4: Integrals 7.2
4. Integrate \sin x\,\sin(\cos x)
Solution
This integral is not immediately of the standard forms \displaystyle \int f(x)f'(x)\,dx or \displaystyle \int \frac{f'(x)}{f(x)}\,dx. Therefore, we look carefully for a suitable substitution.
If we choose t=\sin x, then dt=\cos x\,dx, but no factor of \cos x is present in the integral. Hence, this substitution is not useful.
Similarly, if we choose t=\sin(\cos x), then
\displaystyle dt=-\sin x\,\cos(\cos x)\,dx
but the factor \cos(\cos x) is not present in the integral. Therefore, this substitution also does not help.
Now observe the expression inside the bracket, namely \cos x. If we let t=\cos x, then
\displaystyle dt=-\sin x\,dx
and the factor \sin x\,dx is already present in the integral. Hence this substitution works perfectly.
Important: The success of substitution depends on choosing the right function. Whenever you consider a substitution t=f(x), immediately check whether its derivative f'(x) is also present in the integral. If f'(x)\,dx is available (possibly up to a constant factor), the substitution is likely to work.
Let
t=\cos x
Then
dt=-\sin x\,dx
or
-dt=\sin x\,dx
Substituting, we get
\displaystyle \int \sin x\,\sin(\cos x)\,dx=-\int \sin t\,dt
Using the standard result \displaystyle \int \sin t\,dt=-\cos t+C, we obtain
\displaystyle -\int \sin t\,dt=\cos t+C
Replacing t by \cos x, we get
\displaystyle \int \sin x\,\sin(\cos x)\,dx=\cos(\cos x)+C
Hence,
\boxed{\displaystyle \int \sin x\,\sin(\cos x)\,dx=\cos(\cos x)+C}
Question 5: Constant of Integration
5. Integrate \sin(ax+b)\cos(ax+b)
Solution
Let \displaystyle I=\int \sin(ax+b)\cos(ax+b)\,dx.
This integral is of the form \displaystyle \int f(x)\,f'(x)\,dx.
Here, f(x)=\sin(ax+b) and \displaystyle f'(x)=a\cos(ax+b). The integrand contains \cos(ax+b), which differs from f'(x) only by the constant factor a. This constant can be easily managed during substitution. Therefore, we substitute the function f(x) by letting t=\sin(ax+b).
Let
t=\sin(ax+b)
Then
dt=a\cos(ax+b)\,dx
or
\displaystyle \cos(ax+b)\,dx=\frac{dt}{a}
Substituting, we get
\displaystyle \int \sin(ax+b)\cos(ax+b)\,dx=\frac{1}{a}\int t\,dt
Using the power rule of integration,
\displaystyle \frac{1}{a}\int t\,dt=\frac{1}{a}\cdot\frac{t^2}{2}+C
\displaystyle =\frac{t^2}{2a}+C
Replacing t by \sin(ax+b), we get
\displaystyle \int \sin(ax+b)\cos(ax+b)\,dx=\frac{\sin^2(ax+b)}{2a}+C
Hence,
\boxed{\displaystyle I=\frac{\sin^2(ax+b)}{2a}+C}
Alternative Method
Using the identity
2\sin A\cos A=\sin 2A
we get
\displaystyle \int \sin(ax+b)\cos(ax+b)\,dx=\frac12\int \sin\big(2(ax+b)\big)\,dx
\displaystyle =\frac12\int \sin(2ax+2b)\,dx
Let
t=2ax+2b
Then
dt=2a\,dx
or
\displaystyle dx=\frac{dt}{2a}
Substituting, we get
\displaystyle \frac12\int \sin(2ax+2b)\,dx=\frac{1}{4a}\int \sin t\,dt
\displaystyle =-\frac{\cos t}{4a}+C
Replacing t by 2ax+2b, we obtain
\displaystyle \int \sin(ax+b)\cos(ax+b)\,dx=-\frac{\cos(2ax+2b)}{4a}+C
Hence,
\boxed{\displaystyle I=-\frac{\cos(2ax+2b)}{4a}+C}
Important: The two methods give answers that appear different:
\displaystyle \frac{\sin^2(ax+b)}{2a}+C
and
\displaystyle -\frac{\cos(2ax+2b)}{4a}+C
To compare them, use the identity
\displaystyle \sin^2\theta=\frac{1-\cos 2\theta}{2}
Substituting \theta=ax+b, we get
\displaystyle \frac{\sin^2(ax+b)}{2a}=\frac{1-\cos(2ax+2b)}{4a}
\displaystyle =\frac{1}{4a}-\frac{\cos(2ax+2b)}{4a}
Notice that the two answers differ only by the constant term \frac{1}{4a}. This difference can be absorbed into the constant of integration. Thus, if one answer is written as \displaystyle -\frac{\cos(2ax+2b)}{4a}+C, the other can be written as \displaystyle -\frac{\cos(2ax+2b)}{4a}+C', where C' is another arbitrary constant.
This is why indefinite integrals are called a family of anti-derivatives. Different methods may produce expressions that look different, but they represent the same family of functions and differ only by a constant.
Question 6: Integrals 7.2
6. Integrate \sqrt{ax+b}
Solution
This integral is not in a standard form. However, the expression inside the square root is a linear function whose derivative is a constant. Therefore, substitution is a natural choice.
Let
t=ax+b
Then
dt=a\,dx
or
\displaystyle dx=\frac{dt}{a}
Substituting, we get
\displaystyle \int \sqrt{ax+b}\,dx=\frac{1}{a}\int \sqrt{t}\,dt
\displaystyle =\frac{1}{a}\int t^{1/2}\,dt
Using the power rule of integration,
\displaystyle \frac{1}{a}\int t^{1/2}\,dt=\frac{1}{a}\cdot\frac{t^{3/2}}{3/2}+C
\displaystyle =\frac{2}{3a}t^{3/2}+C
Replacing t by ax+b, we get
\displaystyle \int \sqrt{ax+b}\,dx=\frac{2}{3a}(ax+b)^{3/2}+C
Hence,
\boxed{\displaystyle \int \sqrt{ax+b}\,dx=\frac{2}{3a}(ax+b)^{3/2}+C}
Question 7: Exercise 7.2
7. Integrate x\sqrt{x+2}
Solution
The expression x+2 appears inside the square root. Whenever an expression occurs inside a root, power, logarithm, exponential or trigonometric function, it is often helpful to substitute the inner expression.
Let
t=x+2
Then
dt=dx
and
x=t-2
Substituting, we get
\displaystyle \int x\sqrt{x+2}\,dx=\int (t-2)\sqrt{t}\,dt
\displaystyle =\int \left(t^{3/2}-2t^{1/2}\right)\,dt
Using the power rule of integration,
\displaystyle =\frac{t^{5/2}}{5/2}-2\cdot\frac{t^{3/2}}{3/2}+C
\displaystyle =\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}+C
Replacing t by x+2, we get
\displaystyle \int x\sqrt{x+2}\,dx=\frac{2}{5}(x+2)^{5/2}-\frac{4}{3}(x+2)^{3/2}+C
Hence,
\boxed{\displaystyle \int x\sqrt{x+2}\,dx=\frac{2}{5}(x+2)^{5/2}-\frac{4}{3}(x+2)^{3/2}+C}
Question 8: Integration by Substitution Method
8. Integrate x\sqrt{1+2x^2}
Solution
The expression 1+2x^2 appears inside the square root. Also, its derivative 4x differs from the factor x in the integrand only by a constant. Therefore, substitution is a natural choice.
Let
t=1+2x^2
Then
dt=4x\,dx
or
\displaystyle x\,dx=\frac{dt}{4}
Substituting,
We get, \displaystyle \int x\sqrt{1+2x^2}\,dx=\frac14\int t^{1/2}\,dt
\displaystyle =\frac14\cdot\frac{t^{3/2}}{3/2}+C
\displaystyle =\frac16\,t^{3/2}+C
Replacing t by 1+2x^2, we obtain
\displaystyle \int x\sqrt{1+2x^2}\,dx=\frac16(1+2x^2)^{3/2}+C
Hence,
\boxed{\displaystyle \int x\sqrt{1+2x^2}\,dx=\frac16(1+2x^2)^{3/2}+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.
Question 9: Integrals 7.2
9. Integrate (4x+2)\sqrt{x^2+x+1}
Solution
Let \displaystyle I=\int (4x+2)\sqrt{x^2+x+1}\,dx
The expression x^2+x+1 appears inside the square root. Its derivative is
\displaystyle \frac{d}{dx}(x^2+x+1)=2x+1
Also
\displaystyle 4x+2=2(2x+1)
Thus, the derivative of the expression inside the square root is present up to a constant factor. Therefore, we use substitution.
Let
t=x^2+x+1
Then
dt=(2x+1)\,dx
Since
(4x+2)\,dx=2\,dt
we obtain, \displaystyle \int (4x+2)\sqrt{x^2+x+1}\,dx=2\int t^{1/2}\,dt
\displaystyle =2\cdot\frac{t^{3/2}}{3/2}+C
\displaystyle =\frac43\,t^{3/2}+C
Replacing t by x^2+x+1, we get
\displaystyle \int (4x+2)\sqrt{x^2+x+1}\,dx=\frac43(x^2+x+1)^{3/2}+C
Hence,
\boxed{\displaystyle I=\frac43(x^2+x+1)^{3/2}+C}
Question 10: Integrals 7.2
10. Integrate \displaystyle \frac{1}{x-\sqrt{x}}
Solution
This integral does not immediately appear to be of a standard type. Let us first simplify the denominator by taking \sqrt{x} as a common factor.
\displaystyle \frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}
Therefore,
\displaystyle \int \frac{1}{x-\sqrt{x}}\,dx=\int \frac{dx}{\sqrt{x}(\sqrt{x}-1)}
Now the expression \sqrt{x}-1 appears in the denominator. Since the derivative of \sqrt{x} contains the factor \frac1{\sqrt{x}}, we use the substitution t=\sqrt{x}-1.
Let
t=\sqrt{x}-1
Then
\displaystyle dt=\frac{1}{2\sqrt{x}}\,dx
or
\displaystyle dx=2\sqrt{x}\,dt
Substituting, we get
\displaystyle \int \frac{dx}{\sqrt{x}(\sqrt{x}-1)}=\int \frac{2\sqrt{x}\,dt}{\sqrt{x}\,t}
\displaystyle =2\int \frac{dt}{t}
Using the standard result \displaystyle \int \frac{dt}{t}=\log|t|+C, we obtain
\displaystyle 2\int \frac{dt}{t}=2\log|t|+C
Replacing t by \sqrt{x}-1, we get
\displaystyle \int \frac{1}{x-\sqrt{x}}\,dx=2\log|\sqrt{x}-1|+C
Hence,
\boxed{\displaystyle \int \frac{1}{x-\sqrt{x}}\,dx=2\log|\sqrt{x}-1|+C}
Tip: Before attempting a substitution, simplify the integrand wherever possible. Use trigonometric identities, rewrite fractions, separate factors, or rearrange terms to reveal a function and its derivative. Once the substitution pattern becomes visible, the integral usually reduces to a standard form.
Question 11: Integrals 7.2
11. Integrate \displaystyle \frac{x}{\sqrt{x+4}},\quad x>0
Solution
The expression x+4 appears inside the square root. Whenever an expression occurs inside a root, power, logarithm, exponential or trigonometric function, it is often helpful to substitute the inner expression.
Let
t=x+4
Then
dt=dx
and
x=t-4
Substituting, we get
\displaystyle \int \frac{x}{\sqrt{x+4}}\,dx=\int \frac{t-4}{\sqrt{t}}\,dt
\displaystyle =\int \left(t^{1/2}-4t^{-1/2}\right)\,dt
Using the power rule of integration,
\displaystyle =\frac{t^{3/2}}{3/2}-4\cdot\frac{t^{1/2}}{1/2}+C
\displaystyle =\frac{2}{3}t^{3/2}-8t^{1/2}+C
Replacing t by x+4, we get
\displaystyle \int \frac{x}{\sqrt{x+4}}\,dx=\frac{2}{3}(x+4)^{3/2}-8\sqrt{x+4}+C
Hence,
\displaystyle \int \frac{x}{\sqrt{x+4}}\,dx=\frac{2}{3}(x+4)^{3/2}-8(x+4)^{1/2}+C
Taking \displaystyle \frac{2}{3}\sqrt{x+4} as a common factor, we get
\displaystyle \int \frac{x}{\sqrt{x+4}}\,dx=\frac{2}{3}\sqrt{x+4}\Big((x+4)-12\Big)+C
\displaystyle =\frac{2}{3}(x-8)\sqrt{x+4}+C
Hence,
\boxed{\displaystyle \int \frac{x}{\sqrt{x+4}}\,dx=\frac{2}{3}(x-8)\sqrt{x+4}+C}
Question 12: Integrals 7.2 Exercise
12. Integrate \displaystyle (x^3-1)^{1/3}x^5
Solution
Let \displaystyle I=\int (x^3-1)^{1/3}x^5\,dx
First rewrite the integrand as
\displaystyle I=\int (x^3-1)^{1/3}x^3\cdot x^2\,dx
This integral is of the form \displaystyle [f(x)]^n\,f'(x). Here, the expression x^3-1 appears inside a power and its derivative 3x^2 differs from the factor x^2 only by a constant. Therefore, we use the substitution t=x^3-1.
Let
t=x^3-1
Then
dt=3x^2\,dx
or
\displaystyle x^2\,dx=\frac{dt}{3}
Also,
\displaystyle x^3=t+1
Thus, x^2\,dx is absorbed into dt, while the remaining factor x^3 is replaced by t+1.
Therefore, \displaystyle I=\frac13\int t^{1/3}(t+1)\,dt
\displaystyle =\frac13\int \left(t^{4/3}+t^{1/3}\right)\,dt
\displaystyle =\frac13\left(\frac{t^{7/3}}{7/3}+\frac{t^{4/3}}{4/3}\right)+C
On simplification,
\displaystyle I=\frac17\,t^{7/3}+\frac14\,t^{4/3}+C
Replacing t by x^3-1, we get
\displaystyle I=\frac17(x^3-1)^{7/3}+\frac14(x^3-1)^{4/3}+C
Hence,
\boxed{\displaystyle I=\frac17(x^3-1)^{7/3}+\frac14(x^3-1)^{4/3}+C}
Question 13: Integrals 7.2
13. Integrate \displaystyle \frac{x^2}{(2+3x^3)^3}
Solution
This integral is of the form \displaystyle \frac{f'(x)}{[f(x)]^n}.
Here, f(x)=2+3x^3 and
\displaystyle f'(x)=9x^2
The numerator x^2 differs from f'(x) only by a constant factor. Therefore, we substitute the function f(x) by letting t=2+3x^3.
Let
t=2+3x^3
Then
dt=9x^2\,dx
or
\displaystyle x^2\,dx=\frac{dt}{9}
Substituting, we get
\displaystyle \int \frac{x^2}{(2+3x^3)^3}\,dx=\frac19\int t^{-3}\,dt
Using the power rule of integration,
\displaystyle \frac19\int t^{-3}\,dt=\frac19\cdot\frac{t^{-2}}{-2}+C
\displaystyle =-\frac1{18}t^{-2}+C
Replacing t by 2+3x^3, we get
\displaystyle \int \frac{x^2}{(2+3x^3)^3}\,dx=-\frac1{18(2+3x^3)^2}+C
Hence,
\boxed{\displaystyle \int \frac{x^2}{(2+3x^3)^3}\,dx=-\frac1{18(2+3x^3)^2}+C}
Question 14: Integrals Exercise 7.2
14. Integrate \displaystyle \frac{1}{x(\log x)^m},\quad x>0,\; m\neq 1
Solution
This integral is of the form \displaystyle \frac{f'(x)}{[f(x)]^n}.
Here, f(x)=\log x and
\displaystyle f'(x)=\frac1x
Since the numerator is the derivative of \log x, we substitute the function f(x) by letting t=\log x.
Let
t=\log x
Then
\displaystyle dt=\frac1x\,dx
Substituting, we get
\displaystyle \int \frac{1}{x(\log x)^m}\,dx=\int t^{-m}\,dt
Using the power rule of integration,
\displaystyle \int t^{-m}\,dt=\frac{t^{-m+1}}{-m+1}+C
\displaystyle =\frac{t^{\,1-m}}{1-m}+C\qquad (m\neq 1)
Replacing t by \log x, we get
\displaystyle \int \frac{1}{x(\log x)^m}\,dx=\frac{(\log x)^{\,1-m}}{1-m}+C
Hence,
\boxed{\displaystyle \int \frac{1}{x(\log x)^m}\,dx=\frac{(\log x)^{\,1-m}}{1-m}+C,\quad m\neq 1}
Interesting Fact: While differentiation measures the rate at which a quantity changes, integration helps us recover the original quantity from its rate of change. Together, these two processes form the foundation of calculus.
Question 15: Indefinite Integrals 7.2
15. Integrate \displaystyle \frac{x}{9-4x^2}
Solution
This integral is of the form \displaystyle \frac{f'(x)}{f(x)}.
Here, f(x)=9-4x^2 and
\displaystyle f'(x)=-8x
The numerator x differs from f'(x) only by a constant factor. Therefore, we substitute the function f(x) by letting t=9-4x^2.
Let
t=9-4x^2
Then
dt=-8x\,dx
or
\displaystyle x\,dx=-\frac{dt}{8}
Substituting, we get
\displaystyle \int \frac{x}{9-4x^2}\,dx=-\frac18\int \frac{dt}{t}
Using the standard result \displaystyle \int \frac{dt}{t}=\log|t|+C, we obtain
\displaystyle -\frac18\int \frac{dt}{t}=-\frac18\log|t|+C
Replacing t by 9-4x^2, we get
\displaystyle \int \frac{x}{9-4x^2}\,dx=-\frac18\log|9-4x^2|+C
Hence,
\boxed{\displaystyle \int \frac{x}{9-4x^2}\,dx=-\frac18\log|9-4x^2|+C}
Interesting Fact: Integration is often used to find the total quantity from its rate of change. For example, economists use integration to obtain total cost and total revenue functions from marginal cost and marginal revenue.
Question 16: Integrals 7.2
16. Integrate \displaystyle e^{2x+3}
Solution
The expression 2x+3 appears in the exponent. Whenever an expression occurs inside an exponential, logarithmic, trigonometric function, root or power, it is often helpful to substitute the inner expression.
Let
t=2x+3
Then
dt=2\,dx
or
\displaystyle dx=\frac{dt}{2}
Notice that the derivative of 2x+3 is the constant 2. Therefore, the required factor can be obtained simply by adjusting dx.
Substituting, we get
\displaystyle \int e^{2x+3}\,dx=\frac12\int e^t\,dt
\displaystyle =\frac12 e^t+C
Replacing t by 2x+3, we get
\displaystyle \int e^{2x+3}\,dx=\frac12 e^{2x+3}+C
Hence,
\boxed{\displaystyle \int e^{2x+3}\,dx=\frac12 e^{2x+3}+C}
Question 17: Basic Integrals 7.2
17. Integrate \displaystyle \frac{x}{e^{x^2}}
Solution
First rewrite the integrand as
\displaystyle \frac{x}{e^{x^2}}=xe^{-x^2}
Now the expression x^2 appears in the exponent. Also, its derivative 2x differs from the factor x only by a constant. Therefore, we use the substitution t=-x^2.
Let
t=-x^2
Then
dt=-2x\,dx
or
\displaystyle x\,dx=-\frac{dt}{2}
Substituting, we get
\displaystyle \int xe^{-x^2}\,dx=-\frac12\int e^t\,dt
\displaystyle =-\frac12e^t+C
Replacing t by -x^2, we get
\displaystyle \int \frac{x}{e^{x^2}}\,dx=-\frac12e^{-x^2}+C
Hence,
\boxed{\displaystyle \int \frac{x}{e^{x^2}}\,dx=-\frac12e^{-x^2}+C}
Question 18: Integration by Substitution
18. Integrate \displaystyle \frac{e^{\tan^{-1}x}}{1+x^2}
Solution
The expression \tan^{-1}x appears in the exponent. Also, its derivative
\displaystyle \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}
is already present in the integrand. Therefore, we use the substitution t=\tan^{-1}x.
Let
t=\tan^{-1}x
Then
\displaystyle dt=\frac{dx}{1+x^2}
Substituting, we get
\displaystyle \int \frac{e^{\tan^{-1}x}}{1+x^2}\,dx=\int e^t\,dt
\displaystyle =e^t+C
Replacing t by \tan^{-1}x, we get
\displaystyle \int \frac{e^{\tan^{-1}x}}{1+x^2}\,dx=e^{\tan^{-1}x}+C
Hence,
\boxed{\displaystyle \int \frac{e^{\tan^{-1}x}}{1+x^2}\,dx=e^{\tan^{-1}x}+C}
Question 19: Simplify the Integrand
19. Integrate \displaystyle \frac{e^{2x}-1}{e^{2x}+1}
Solution
At first sight, the integral does not appear to be of a standard type. In such situations, look for a way to rewrite the expression so that a suitable substitution becomes visible.
Since both the numerator and denominator contain powers of e^x, we divide them by e^x. This gives
\displaystyle \frac{e^{2x}-1}{e^{2x}+1}=\frac{e^x-e^{-x}}{e^x+e^{-x}}
Therefore,
\displaystyle I=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx
Now notice that
\displaystyle \frac{d}{dx}(e^x+e^{-x})=e^x-e^{-x}
Thus, the integral has been reduced to the standard form \displaystyle \frac{f'(x)}{f(x)}.
Let
t=e^x+e^{-x}
Then
dt=(e^x-e^{-x})\,dx
Substituting, we get
\displaystyle I=\int \frac{dt}{t}
\displaystyle =\log|t|+C
Replacing t by e^x+e^{-x}, we obtain
\displaystyle I=\log|e^x+e^{-x}|+C
Hence,
\boxed{\displaystyle \int \frac{e^{2x}-1}{e^{2x}+1}\,dx=\log|e^x+e^{-x}|+C}
Important: A suitable substitution is not always visible at first glance. When the integrand does not resemble a familiar pattern, try simplifying or rewriting it. In many questions, the correct substitution becomes apparent only after the expression has been rearranged. In Q. 19 of Integrals 7.2, dividing the numerator and denominator by e^x transforms the integrand into the standard form \frac{f'(x)}{f(x)}, making the substitution straightforward.
Question 20: Integrals 7.2
20. Integrate \displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}
Solution
Unlike Q.19, this integral is already of the form \displaystyle \frac{f'(x)}{f(x)}.
Here,
f(x)=e^{2x}+e^{-2x}
and
\displaystyle f'(x)=2e^{2x}-2e^{-2x}=2\left(e^{2x}-e^{-2x}\right)
The numerator differs from f'(x) only by the constant factor 2. Therefore, we substitute the function f(x) by letting t=e^{2x}+e^{-2x}.
Let
t=e^{2x}+e^{-2x}
Then
dt=2\left(e^{2x}-e^{-2x}\right)\,dx
or
\displaystyle \left(e^{2x}-e^{-2x}\right)\,dx=\frac{dt}{2}
Substituting, we get
\displaystyle \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx=\frac12\int \frac{dt}{t}
\displaystyle =\frac12\log|t|+C
Replacing t by e^{2x}+e^{-2x}, we get
\displaystyle \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx=\frac12\log|e^{2x}+e^{-2x}|+C
Hence,
\boxed{\displaystyle \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx=\frac12\log|e^{2x}+e^{-2x}|+C}
Question 21: Trigonometric Integrals
21. Integrate \displaystyle \tan^2(2x-3)
Solution
The integrand contains \tan^2(2x-3). To simplify it, we use the identity
\displaystyle \tan^2\theta=\sec^2\theta-1
Therefore,
\displaystyle \int \tan^2(2x-3)\,dx=\int \Big(\sec^2(2x-3)-1\Big)\,dx
\displaystyle =\int \sec^2(2x-3)\,dx-\int dx
For the first integral, let
t=2x-3
Then
dt=2\,dx
or
\displaystyle dx=\frac{dt}{2}
Hence, \displaystyle \int \sec^2(2x-3)\,dx=\frac12\int \sec^2 t\,dt
\displaystyle =\frac12\tan t+C
\displaystyle =\frac12\tan(2x-3)+C
Therefore,
\displaystyle \int \tan^2(2x-3)\,dx=\frac12\tan(2x-3)-x+C
Hence,
\boxed{\displaystyle \int \tan^2(2x-3)\,dx=\frac12\tan(2x-3)-x+C}
Question 22: Trigonometric Integrals
22. Integrate \displaystyle \sec^2(7-4x)
Solution
The expression 7-4x appears inside the trigonometric function. Since the derivative of 7-4x is the constant -4, substitution is a natural choice.
Let
t=7-4x
Then
dt=-4\,dx
or
\displaystyle dx=-\frac{dt}{4}
Substituting, we get
\displaystyle \int \sec^2(7-4x)\,dx=-\frac14\int \sec^2 t\,dt
Using the standard result
\displaystyle \int \sec^2 t\,dt=\tan t+C
we obtain
\displaystyle -\frac14\int \sec^2 t\,dt=-\frac14\tan t+C
Replacing t by 7-4x, we get
\displaystyle \int \sec^2(7-4x)\,dx=-\frac14\tan(7-4x)+C
Hence,
\boxed{\displaystyle \int \sec^2(7-4x)\,dx=-\frac14\tan(7-4x)+C}
Shortcut: If the entire integrand is of the form f(ax+b), then
\displaystyle \int f(ax+b)\,dx=\frac1a\,F(ax+b)+C
where F'(x)=f(x). In simple words, find the usual integral and then divide the answer by the derivative of ax+b.
For example, \displaystyle \int \sec^2(2x)\,dx=\frac{\tan(2x)}2+C and \displaystyle \int e^{5x-2}\,dx=\frac{e^{5x-2}}5+C.
This shortcut does not directly evaluate integrals such as \displaystyle \int x\sin(3x+1)\,dx, since the entire integrand is not of the form f(ax+b). However, it can still be applied to the \sin(3x+1) part when using other methods such as integration by parts.
Question 23: Integrals 7.2
23. Integrate \displaystyle \frac{\sin^{-1}x}{\sqrt{1-x^2}}
Solution
The function \sin^{-1}x appears in the numerator. Also, its derivative is
\displaystyle \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}
which is already present in the integrand. Therefore, the integral is of the form \displaystyle f(x)f'(x), and we use the substitution t=\sin^{-1}x.
Let
t=\sin^{-1}x
Then
\displaystyle dt=\frac{dx}{\sqrt{1-x^2}}
Substituting, we get
\displaystyle \int \frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx=\int t\,dt
Using the power rule of integration,
\displaystyle \int t\,dt=\frac{t^2}{2}+C
Replacing t by \sin^{-1}x, we get
\displaystyle \int \frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx=\frac{(\sin^{-1}x)^2}{2}+C
Hence,
\boxed{\displaystyle \int \frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx=\frac{(\sin^{-1}x)^2}{2}+C}
Question 24: Indefinite Integrals 7.2
24. Integrate \displaystyle \frac{2\cos x-3\sin x}{6\cos x+4\sin x}
Solution
Let \displaystyle I=\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx
First take 2 common from the denominator, we get
\displaystyle I=\frac12\int \frac{2\cos x-3\sin x}{3\cos x+2\sin x}\,dx
Now the integral is of the exact form \displaystyle \frac{f'(x)}{f(x)}.
Here,
f(x)=3\cos x+2\sin x
and
\displaystyle f'(x)=-3\sin x+2\cos x=2\cos x-3\sin x
Thus, the numerator is the exact derivative of the denominator. Therefore, we use the substitution
Let
t=3\cos x+2\sin x
Then
\displaystyle dt=(2\cos x-3\sin x)\,dx
Substituting,
We get, \displaystyle \frac12\int \frac{dt}{t}
\displaystyle =\frac12\log|t|+C
\displaystyle =\frac12\log|3\cos x+2\sin x|+C
Hence,
\boxed{\displaystyle I=\frac12\log|3\cos x+2\sin x|+C}
Question 25: Integrals 7.2
25. Integrate \displaystyle \frac{1}{\cos^2x\,(1-\tan x)^2}
Solution
First rewrite the integrand using the identity
\displaystyle \sec^2x=\frac{1}{\cos^2x}
Then
\displaystyle \int \frac{1}{\cos^2x\,(1-\tan x)^2}\,dx=\int \frac{\sec^2x}{(1-\tan x)^2}\,dx
Now the integral is of the form \displaystyle \frac{f'(x)}{[f(x)]^n}.
Here,
f(x)=1-\tan x
and
\displaystyle f'(x)=-\sec^2x
The numerator differs from f'(x) only by a constant factor. Therefore, we use the substitution
t=1-\tan x
Then
\displaystyle dt=-\sec^2x\,dx
or
\displaystyle \sec^2x\,dx=-dt
Substituting, we get
\displaystyle \int \frac{\sec^2x}{(1-\tan x)^2}\,dx=-\int t^{-2}\,dt
Using the power rule of integration,
\displaystyle -\int t^{-2}\,dt=-\left(\frac{t^{-1}}{-1}\right)+C
\displaystyle =\frac{1}{t}+C
Replacing t by 1-\tan x, we get
\displaystyle \int \frac{1}{\cos^2x\,(1-\tan x)^2}\,dx=\frac{1}{1-\tan x}+C
Hence,
\boxed{\displaystyle \int \frac{1}{\cos^2x\,(1-\tan x)^2}\,dx=\frac{1}{1-\tan x}+C}
Question 26: Integration by Substitution
26. Integrate \displaystyle \frac{\cos\sqrt{x}}{\sqrt{x}}
Solution
The expression \sqrt{x} appears inside the trigonometric function. Also, its derivative contains the factor \frac1{\sqrt{x}}, which is already present in the integrand. Therefore, we use the substitution t=\sqrt{x}.
Let
t=\sqrt{x}
Then
\displaystyle dt=\frac{1}{2\sqrt{x}}\,dx
or
\displaystyle \frac{dx}{\sqrt{x}}=2\,dt
Substituting, we get
\displaystyle \int \frac{\cos\sqrt{x}}{\sqrt{x}}\,dx=2\int \cos t\,dt
\displaystyle =2\sin t+C
Replacing t by \sqrt{x}, we get
\displaystyle \int \frac{\cos\sqrt{x}}{\sqrt{x}}\,dx=2\sin\sqrt{x}+C
Hence,
\boxed{\displaystyle \int \frac{\cos\sqrt{x}}{\sqrt{x}}\,dx=2\sin\sqrt{x}+C}
Question 27: Integrals 7.2
27. Integrate \displaystyle \sqrt{\sin 2x}\,\cos 2x
Solution
The expression \sin 2x appears inside the square root. Also, its derivative
\displaystyle \frac{d}{dx}(\sin 2x)=2\cos 2x
differs from the factor \cos 2x only by a constant. Therefore, we use the substitution t=\sin 2x.
Let
t=\sin 2x
Then
\displaystyle dt=2\cos 2x\,dx
or
\displaystyle \cos 2x\,dx=\frac{dt}{2}
Substituting, we get
\displaystyle \int \sqrt{\sin 2x}\,\cos 2x\,dx=\frac12\int t^{1/2}\,dt
Using the power rule of integration,
\displaystyle \frac12\int t^{1/2}\,dt=\frac12\cdot\frac{t^{3/2}}{3/2}+C
\displaystyle =\frac13\,t^{3/2}+C
Replacing t by \sin 2x, we get
\displaystyle \int \sqrt{\sin 2x}\,\cos 2x\,dx=\frac13(\sin 2x)^{3/2}+C
Hence,
\boxed{\displaystyle \int \sqrt{\sin 2x}\,\cos 2x\,dx=\frac13(\sin 2x)^{3/2}+C}
Observation: Questions 26 and 27 of Integrals 7.2 show a common substitution pattern. Whenever an expression appears inside a trigonometric function and its derivative (or a constant multiple of it) is present in the integrand, substitution usually converts the integral into a standard trigonometric integral.
Question 28: Integrals Exercise 7.2
28. Integrate \displaystyle \frac{\cos x}{\sqrt{1+\sin x}}
Solution
The expression 1+\sin x appears inside the square root. Also, its derivative
\displaystyle \frac{d}{dx}(1+\sin x)=\cos x
is already present in the numerator. Therefore, we use the substitution t=1+\sin x.
Let
t=1+\sin x
Then
\displaystyle dt=\cos x\,dx
Substituting, we get
\displaystyle \int \frac{\cos x}{\sqrt{1+\sin x}}\,dx=\int t^{-1/2}\,dt
Using the power rule of integration,
\displaystyle \int t^{-1/2}\,dt=\frac{t^{1/2}}{1/2}+C
\displaystyle =2\sqrt{t}+C
Replacing t by 1+\sin x, we get
\displaystyle \int \frac{\cos x}{\sqrt{1+\sin x}}\,dx=2\sqrt{1+\sin x}+C
Hence,
\boxed{\displaystyle \int \frac{\cos x}{\sqrt{1+\sin x}}\,dx=2\sqrt{1+\sin x}+C}
Question 29: Trigonometric Integrals
29. Integrate \displaystyle \cot x\,\log(\sin x)
Solution
The function \log(\sin x) appears in the integrand. Also,
\displaystyle \frac{d}{dx}\big(\log(\sin x)\big)=\frac{\cos x}{\sin x}=\cot x
which is already present in the integrand. Therefore, the integral is of the form \displaystyle f(x)f'(x), and we use the substitution t=\log(\sin x).
Let
t=\log(\sin x)
Then
\displaystyle dt=\cot x\,dx
Substituting, we get
\displaystyle \int \cot x\,\log(\sin x)\,dx=\int t\,dt
Using the power rule of integration,
\displaystyle \int t\,dt=\frac{t^2}{2}+C
Replacing t by \log(\sin x), we get
\displaystyle \int \cot x\,\log(\sin x)\,dx=\frac12\big(\log(\sin x)\big)^2+C
Hence,
\boxed{\displaystyle \int \cot x\,\log(\sin x)\,dx=\frac12\big(\log(\sin x)\big)^2+C}
Question 30: Indefinite Integrals 7.2
30. Integrate \displaystyle \frac{\sin x}{1+\cos x}
Solution
This integral is of the form \displaystyle \frac{f'(x)}{f(x)}.
Here,
f(x)=1+\cos x
and
\displaystyle f'(x)=-\sin x
The numerator differs from f'(x) only by a negative sign. Therefore, we use the substitution
t=1+\cos x
Then
\displaystyle dt=-\sin x\,dx
or
\displaystyle \sin x\,dx=-dt
Substituting, we get
\displaystyle \int \frac{\sin x}{1+\cos x}\,dx=-\int \frac{dt}{t}
\displaystyle =-\log|t|+C
Replacing t by 1+\cos x, we obtain
\displaystyle \int \frac{\sin x}{1+\cos x}\,dx=-\log|1+\cos x|+C
Hence,
\boxed{\displaystyle \int \frac{\sin x}{1+\cos x}\,dx=-\log|1+\cos x|+C}
Question 31: Integration by Substitution
31. Integrate \displaystyle \frac{\sin x}{(1+\cos x)^2}
Solution
Similar to Q. 30, this integral is of the form \displaystyle \frac{f'(x)}{[f(x)]^n}.
Here,
f(x)=1+\cos x
and
\displaystyle f'(x)=-\sin x
The numerator differs from f'(x) only by a negative sign. Therefore, we use the substitution
t=1+\cos x
Then
\displaystyle dt=-\sin x\,dx
or
\displaystyle \sin x\,dx=-dt
Substituting, we get
\displaystyle \int \frac{\sin x}{(1+\cos x)^2}\,dx=-\int t^{-2}\,dt
Using the power rule of integration,
\displaystyle -\int t^{-2}\,dt=-\left(\frac{t^{-1}}{-1}\right)+C
\displaystyle =\frac1t+C
Replacing t by 1+\cos x, we get
\displaystyle \int \frac{\sin x}{(1+\cos x)^2}\,dx=\frac{1}{1+\cos x}+C
Hence,
\boxed{\displaystyle \int \frac{\sin x}{(1+\cos x)^2}\,dx=\frac{1}{1+\cos x}+C}
Tip: Questions such as Q. 32, Q. 33 and Q. 34 may not look suitable for substitution at first sight. In such cases, try rewriting the trigonometric expressions and look for a hidden \frac{f'(x)}{f(x)} or f'(x)[f(x)]^n pattern. These special forms are worth remembering as they appear frequently in examinations.
Question 32: Special Integrals 7.2
32. Integrate \displaystyle \frac{1}{1+\cot x}
Solution
Remember in this question, we will first express \cot x in terms of \sin x and \cos x.
\displaystyle \int \frac{1}{1+\cot x}\,dx=\int \frac{1}{1+\frac{\cos x}{\sin x}}\,dx
\displaystyle =\int \frac{\sin x}{\sin x+\cos x}\,dx
Now write \sin x as (remember this important step):
\displaystyle \sin x=\frac12(2\sin x)
\displaystyle \sin x=\frac12(\sin x+\sin x)
Now adding and subtracting \cos x, which will not change the value, we get
\displaystyle \sin x=\frac12\Big[(\sin x+\cos x)+(\sin x-\cos x)\Big]
Therefore,
\displaystyle \frac{\sin x}{\sin x+\cos x}=\frac12+\frac12\cdot\frac{\sin x-\cos x}{\sin x+\cos x}
Hence,
\displaystyle I=\frac12\int dx+\frac12\int \frac{\sin x-\cos x}{\sin x+\cos x}\,dx
For the second integral, let
t=\sin x+\cos x
Then
\displaystyle dt=(\cos x-\sin x)\,dx
or
\displaystyle (\sin x-\cos x)\,dx=-dt
Substituting, we get
\displaystyle I=\frac{x}{2}-\frac12\int \frac{dt}{t}
\displaystyle =\frac{x}{2}-\frac12\log|t|+C
Replacing t by \sin x+\cos x, we obtain
\displaystyle I=\frac{x}{2}-\frac12\log|\sin x+\cos x|+C
Hence,
\boxed{\displaystyle \int \frac{1}{1+\cot x}\,dx=\frac{x}{2}-\frac12\log|\sin x+\cos x|+C}
Question 33: Special Integrals 7.2
33. Integrate \displaystyle \frac{1}{1-\tan x}
Solution
Just like in Q. 32, first express \tan x in terms of \sin x and \cos x.
\displaystyle \int \frac{1}{1-\tan x}\,dx=\int \frac{1}{1-\frac{\sin x}{\cos x}}\,dx
\displaystyle =\int \frac{\cos x}{\cos x-\sin x}\,dx
Now write \cos x as (as we wrote \sin x in Q. 32
\displaystyle \cos x=\frac12\Big[(\cos x-\sin x)+(\cos x+\sin x)\Big]
Therefore,
\displaystyle \frac{\cos x}{\cos x-\sin x}=\frac12+\frac12\cdot\frac{\cos x+\sin x}{\cos x-\sin x}
Hence,
\displaystyle I=\frac12\int dx+\frac12\int \frac{\cos x+\sin x}{\cos x-\sin x}\,dx
For the second integral, let
t=\cos x-\sin x
Then
\displaystyle dt=-(\sin x+\cos x)\,dx
or
\displaystyle (\sin x+\cos x)\,dx=-dt
Substituting, we get
\displaystyle I=\frac{x}{2}-\frac12\int \frac{dt}{t}
\displaystyle =\frac{x}{2}-\frac12\log|t|+C
Replacing t by \cos x-\sin x, we obtain
\displaystyle I=\frac{x}{2}-\frac12\log|\cos x-\sin x|+C
Hence,
\boxed{\displaystyle \int \frac{1}{1-\tan x}\,dx=\frac{x}{2}-\frac12\log|\cos x-\sin x|+C}
Question 34: Integrals 7.2
34. Integrate \displaystyle \frac{\sqrt{\tan x}}{\sin x\cos x}
Solution
First rewrite the integrand in a form suitable for substitution.
We have, \displaystyle \frac{\sqrt{\tan x}}{\sin x\cos x}=\sqrt{\tan x}\cdot\frac{1}{\cos^2x}\cdot\frac{\cos x}{\sin x}
\displaystyle =\sqrt{\tan x}\cdot\sec^2x\cdot\frac1{\tan x}
\displaystyle =\frac{\sec^2x}{\sqrt{\tan x}}
Therefore, \displaystyle \int \frac{\sqrt{\tan x}}{\sin x\cos x}\,dx=\int \frac{\sec^2x}{\sqrt{\tan x}\,dx
Now the integral is of the form \displaystyle \frac{f'(x)}{\sqrt{f(x)}}.
Let
t=\tan x
Then
\displaystyle dt=\sec^2x\,dx
Substituting,
We get, \displaystyle \int \frac{\sec^2x}{\sqrt{\tan x}\,dx=\int t^{-1/2}\,dt
\displaystyle =\frac{t^{1/2}}{1/2}+C
\displaystyle =2\sqrt{t}+C
Replacing t by \tan x, we obtain
\displaystyle \int \frac{\sqrt{\tan x}}{\sin x\cos x}\,dx=2\sqrt{\tan x}+C
Hence,
\boxed{\displaystyle \int \frac{\sqrt{\tan x}}{\sin x\cos x}\,dx=2\sqrt{\tan x}+C}
Question 35: Integration by Substitution
35. Integrate \displaystyle \frac{(1+\log x)^2}{x}
Solution
The expression 1+\log x appears inside a power. Also, its derivative
\displaystyle \frac{d}{dx}(1+\log x)=\frac1x
is already present in the integrand. Therefore, the integral is of the form \displaystyle [f(x)]^n\,f'(x), and we use the substitution t=1+\log x.
Let
t=1+\log x
Then
\displaystyle dt=\frac{dx}{x}
Substituting, we get
\displaystyle \int \frac{(1+\log x)^2}{x}\,dx=\int t^2\,dt
Using the power rule of integration,
\displaystyle \int t^2\,dt=\frac{t^3}{3}+C
Replacing t by 1+\log x, we get
\displaystyle \int \frac{(1+\log x)^2}{x}\,dx=\frac{(1+\log x)^3}{3}+C
Hence,
\boxed{\displaystyle \int \frac{(1+\log x)^2}{x}\,dx=\frac{(1+\log x)^3}{3}+C}
Observation: As the exercise progresses, the challenge shifts from performing the substitution to identifying it. If the correct substitution is found, most integrals reduce to a standard form. Therefore, spend a few moments analyzing the structure of the integrand before starting the calculations.
Question 36: Simplify the Integrand
36. Integrate \displaystyle \frac{(x+1)(x+\log x)^2}{x}
Solution
First rewrite the integrand as
\displaystyle \frac{(x+1)(x+\log x)^2}{x}=\left(1+\frac1x\right)(x+\log x)^2
Now observe that
\displaystyle \frac{d}{dx}(x+\log x)=1+\frac1x
which is already present as a factor in the integrand. Therefore, the integral is of the form \displaystyle [f(x)]^n\,f'(x).
Let
t=x+\log x
Then
\displaystyle dt=\left(1+\frac1x\right)dx
Substituting, we get
\displaystyle \int \frac{(x+1)(x+\log x)^2}{x}\,dx=\int t^2\,dt
Using the power rule of integration,
\displaystyle \int t^2\,dt=\frac{t^3}{3}+C
Replacing t by x+\log x, we get
\displaystyle \int \frac{(x+1)(x+\log x)^2}{x}\,dx=\frac{(x+\log x)^3}{3}+C
Hence,
\boxed{\displaystyle \int \frac{(x+1)(x+\log x)^2}{x}\,dx=\frac{(x+\log x)^3}{3}+C}
Question 37: Integration by Substitution
37. Integrate \displaystyle \frac{x^3\sin(\tan^{-1}x^4)}{1+x^8}
Solution
The expression \tan^{-1}x^4 appears inside the trigonometric function. Also,
\displaystyle \frac{d}{dx}\left(\tan^{-1}x^4\right)=\frac{4x^3}{1+x^8}
which differs from the factor \displaystyle \frac{x^3}{1+x^8} only by a constant. Therefore, we use the substitution
t=\tan^{-1}x^4
Then
\displaystyle dt=\frac{4x^3}{1+x^8}\,dx
or
\displaystyle \frac{x^3}{1+x^8}\,dx=\frac{dt}{4}
Substituting, we get
\displaystyle \int \frac{x^3\sin(\tan^{-1}x^4)}{1+x^8}\,dx=\frac14\int \sin t\,dt
\displaystyle =-\frac14\cos t+C
Replacing t by \tan^{-1}x^4, we get
\displaystyle \int \frac{x^3\sin(\tan^{-1}x^4)}{1+x^8}\,dx=-\frac14\cos(\tan^{-1}x^4)+C
Hence,
\boxed{\displaystyle \int \frac{x^3\sin(\tan^{-1}x^4)}{1+x^8}\,dx=-\frac14\cos(\tan^{-1}x^4)+C}
Question 38: Integrals 7.2 – MCQ
38. Choose the correct answer:
\displaystyle \int \frac{10x^9+10^x\log_e 10}{x^{10}+10^x}\,dx
- (A) 10^x-x^{10}+C
- (B) 10^x+x^{10}+C
- (C) (10^x-x^{10})^{-1}+C
- (D) \log(10^x+x^{10})+C
Solution
Let
f(x)=x^{10}+10^x
Then
\displaystyle f'(x)=10x^9+10^x\log_e 10
which is exactly the numerator.
Therefore, the integral is of the form \displaystyle \frac{f'(x)}{f(x)}.
Using the standard result
\displaystyle \int \frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C
we get
\displaystyle \int \frac{10x^9+10^x\log_e 10}{x^{10}+10^x}\,dx=\log(x^{10}+10^x)+C
✅️ Hence, the correct answer is (D).
Question 39: Integrals 7.2 – MCQ
39. Choose the correct answer:
\displaystyle \int \frac{dx}{\sin^2x\cos^2x}
- (A) \tan x+\cot x+C
- (B) \tan x-\cot x+C
- (C) \tan x\cot x+C
- (D) \tan x-\cot 2x+C
Solution
Using the identities
\displaystyle \sec^2x=\frac1{\cos^2x}\qquad\text{and}\qquad \cosec^2x=\frac1{\sin^2x}
we can write
\displaystyle \frac1{\sin^2x\cos^2x}=\sec^2x+\cosec^2x
since
\displaystyle \sec^2x+\cosec^2x=\frac1{\cos^2x}+\frac1{\sin^2x}=\frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\frac1{\sin^2x\cos^2x}
Therefore, \displaystyle \int \frac{dx}{\sin^2x\cos^2x}=\int(\sec^2x+\cosec^2x)\,dx
\displaystyle =\int \sec^2x\,dx+\int \cosec^2x\,dx
\displaystyle =\tan x-\cot x+C
✅️ Hence, the correct answer is (B).
Common Mistakes to Avoid
- Choosing the wrong substitution: The substitution should usually be the expression whose derivative (or a constant multiple of it) is also present in the integrand.
- Not looking for standard patterns: Before starting the calculations, check whether the integral is of the form \displaystyle f(x)f'(x), \displaystyle \frac{f'(x)}{f(x)}, \displaystyle [f(x)]^n f'(x) or \displaystyle \frac{f'(x)}{[f(x)]^n}.
- Ignoring constant factors: The derivative need not match exactly. If it differs only by a constant factor, adjust the integral accordingly instead of abandoning the substitution.
- Forgetting to change dx: After substituting t=f(x), always replace dx using dt=f'(x)\,dx.
- Not rewriting the integrand first: Many questions become easier after simplifying fractions, using trigonometric identities, or rearranging factors to reveal a suitable substitution.
- Leaving x and t together: After substitution, the integral should be written entirely in terms of the new variable.
- Forgetting to substitute back: After integrating with respect to t, replace t by the original expression before writing the final answer.
- Missing hidden derivatives: Expressions such as \tan^{-1}(x^4), \log(\sin x) or \sqrt{x} often contain inner functions whose derivatives are hidden elsewhere in the integrand.
- Forgetting the constant of integration: Every indefinite integral must include 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 7.2 of Integrals, you should now be familiar with the method of substitution, one of the most important techniques of integration. You have learned how to identify suitable substitutions, transform the integral into a simpler form, and evaluate a wide variety of integrals involving algebraic, trigonometric, logarithmic and exponential functions.
To strengthen your understanding further, make sure that you revise:
- The idea behind substitution and why it simplifies an integral
- Identifying a suitable substitution by looking for a function and its derivative in the integrand
- Standard patterns such as \displaystyle f(x)f'(x), \displaystyle \frac{f'(x)}{f(x)}, \displaystyle [f(x)]^n f'(x) and \displaystyle \frac{f'(x)}{[f(x)]^n}
- Handling cases where the derivative differs only by a constant factor
- Changing dx correctly after making a substitution
- Rewriting trigonometric expressions to reveal a suitable substitution
- Simplifying fractions, radicals and algebraic expressions before attempting substitution
- Expressing the entire integral in terms of the new variable before integrating
- Substituting back the original expression after integration
- Checking answers by differentiating the final result whenever possible
Explore More
The substitution method becomes easier with practice. As you work through more questions, focus on identifying a suitable substitution, looking for a function and its derivative within the integrand, and simplifying expressions whenever the required pattern is not immediately visible. Remember that the derivative need not match exactly—it may differ by a constant factor. Finally, after integrating, always substitute back the original expression and include the constant of integration C. With regular practice, you’ll develop the ability to recognize substitution patterns quickly and solve integrals more efficiently.
All the best and keep learning 👍
