Integrals: Chapter 7 Links
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Integrals 7.3 focuses on the evaluation of integrals using trigonometric identities. In many problems, the integrand cannot be integrated directly using standard formulas. In such cases, trigonometric identities help us simplify the expression into a form that is easier to integrate.
\displaystyle \int f(x)\,dx = \int \big(\text{Equivalent Trigonometric Form}\big)\,dx
Some of the most commonly used identities in this exercise include the double-angle identities, triple-angle identities, and product-to-sum or product-to-difference formulas. By applying these identities appropriately, complicated trigonometric expressions can often be converted into simpler forms.
In this exercise, we will learn how to use trigonometric identities in integration and evaluate integrals involving products and powers of trigonometric functions. These techniques are important for solving a wide variety of integration problems that appear in NCERT, CBSE Board Exams, and other competitive examinations.
Key Concepts
1. Integration Using Trigonometric Identities
Many trigonometric expressions cannot be integrated directly using standard integration formulas. In such cases, we first simplify the integrand by applying suitable trigonometric identities and then perform the integration.
Example:
\cos^2x=\frac{1+\cos2x}{2}
Therefore,
\displaystyle \int \cos^2x\,dx=\frac12\int(1+\cos2x)\,dx
This converts the given integral into a form that can be evaluated using standard integration formulas.
2. Common Trigonometric Identities Used
The most frequently used identities in this exercise are:
- Double-angle identities
- Triple-angle identities
- Product-to-sum identities
- Sum-to-product identities
- Power reduction identities
A quick revision table of these identities is provided below for reference.
3. Product-to-Sum Transformations
Products of trigonometric functions are often converted into sums or differences because sums are usually easier to integrate.
Example:
\sin A\cos B=\frac12\big[\sin(A+B)+\sin(A-B)\big]
After applying the identity, each term can be integrated separately.
4. Use of Linearity Property
After simplifying the integrand, we use the linearity property of integration to split the integral into simpler parts.
\displaystyle \int \big(f(x)\pm g(x)\big)\,dx=\int f(x)\,dx\pm\int g(x)\,dx
This property is used repeatedly throughout this exercise.
5. General Strategy for Exercise 7.3
- Identify the trigonometric expression involved.
- Select an appropriate trigonometric identity.
- Simplify the integrand using the identity.
- Split the integral into simpler terms if required.
- Apply standard integration formulas.
- Always add the constant of integration C.
Most questions in this exercise become straightforward once the correct trigonometric identity is identified and applied.
Tip: Most integrals in this exercise cannot be evaluated directly. First identify a suitable trigonometric identity to simplify the integrand, then apply the standard rules of integration. After finding the anti-derivative, always include the constant of integration C.
Important Trigonometric Identities
The following trigonometric identities are frequently used in this exercise. Before integrating, many expressions need to be transformed using double-angle, triple-angle, product-to-sum, sum-to-product, or power-reduction formulas. Once the integrand is simplified, the required integral can usually be evaluated using standard integration formulas.
| Identity Type | Formula |
|---|---|
| Angle Addition & Subtraction | \sin(A\pm B)=\sin A\cos B\pm\cos A\sin B |
| Angle Addition & Subtraction | \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B |
| Double-Angle (2A) | \sin2A=2\sin A\cos A=\frac{2\tan A}{1+\tan^2A} |
| Double-Angle (2A) | \cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A=\frac{1-\tan^2A}{1+\tan^2A} |
| Double-Angle (2A) | \tan2A=\frac{2\tan A}{1-\tan^2A} |
| Double-Angle (2A) | \cot2A=\frac{\cot^2A-1}{2\cot A} |
| Triple-Angle (3A) | \sin3A=3\sin A-4\sin^3A |
| Triple-Angle (3A) | \cos3A=4\cos^3A-3\cos A |
| Triple-Angle (3A) | \tan3A=\frac{3\tan A-\tan^3A}{1-3\tan^2A} |
| Cubic Forms | \sin^3A=\frac{3\sin A-\sin3A}{4} |
| Cubic Forms | \cos^3A=\frac{3\cos A+\cos3A}{4} |
| Product to Sum | \sin A\cos B=\frac12\big[\sin(A+B)+\sin(A-B)\big] |
| Product to Difference | \cos A\sin B=\frac12\big[\sin(A+B)-\sin(A-B)\big] |
| Product to Sum | \cos A\cos B=\frac12\big[\cos(A+B)+\cos(A-B)\big] |
| Product to Difference | \sin A\sin B=\frac12\big[\cos(A-B)-\cos(A+B)\big] |
| Sum to Product | \sin C+\sin D=2\sin\frac{C+D}{2}\cos\frac{C-D}{2} |
| Difference to Product | \sin C-\sin D=2\cos\frac{C+D}{2}\sin\frac{C-D}{2} |
| Sum to Product | \cos C+\cos D=2\cos\frac{C+D}{2}\cos\frac{C-D}{2} |
| Differece to Product | \cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2} |
| Power Reduction | \sin^2A=\frac{1-\cos2A}{2} |
| Power Reduction | \cos^2A=\frac{1+\cos2A}{2} |
| Tan² Forms | \tan^2A=\frac{1-\cos2A}{1+\cos2A} |
| Cot² Forms | \cot^2A=\frac{1+\cos2A}{1-\cos2A} |
Useful Convention: While applying Product-to-Sum and Sum-to-Product identities, we shall generally assume that A>B and C>D. This keeps expressions such as A-B and C-D positive, reduces unnecessary simplification steps, and helps avoid sign errors.
Let us now solve all the NCERT questions step by step in this Exercise 7.3 of Integrals.
Question 1: Integrals 7.3
1. Evaluate \int \sin^2(2x+5)\,dx.
Solution
Let, \displaystyle I=\int \sin^2(2x+5)\,dx
Using the power reduction identity
\displaystyle \sin^2A=\frac{1-\cos2A}{2}
Putting A=2x+5, we get
\displaystyle \sin^2(2x+5)=\frac{1-\cos(4x+10)}{2}
Therefore,
\displaystyle I=\int \frac{1-\cos(4x+10)}{2}\,dx
Using the properties of indefinite integrals, we get
\displaystyle I=\frac12\int 1\,dx-\frac12\int \cos(4x+10)\,dx
Now,
\displaystyle \int \cos(4x+10)\,dx=\frac{\sin(4x+10)}{4}
Remember the following standard results:
- \displaystyle \int \sin(ax+b)\,dx=-\frac{\cos(ax+b)}{a}+C
- \displaystyle \int \cos(ax+b)\,dx=\frac{\sin(ax+b)}{a}+C
Note: The above formulas were obtained in Exercise 7.2, Question. 22 using the substitution t=ax+b. We shall use them directly throughout this exercise/chapter and later in Differential Equations frequently.
Therefore,
\displaystyle I=\frac{x}{2}-\frac12\left(\frac{\sin(4x+10)}{4}\right)+C
\displaystyle =\frac{x}{2}-\frac{\sin(4x+10)}{8}+C
Hence,
\boxed{\displaystyle I=\frac{x}{2}-\frac18\sin(4x+10)+C}
Question 2: Integrals Ex. 7.3
2. Evaluate \int \sin 3x\cos 4x\,dx.
Solution
Let, \displaystyle I=\int \sin 3x\cos 4x\,dx
Rearranging the terms, we get
\displaystyle I=\int \cos 4x\sin 3x\,dx
💡We have written \cos4x before \sin3x so that while applying the Product-to-Sum identity, the larger angle can be taken as A and the smaller angle as B. This helps us avoid unnecessary negative signs arising from expressions like \sin(3x-4x)=\sin(-x) and makes the calculations simpler. This step is not necessary and the question can be solved without rearranging the factors as well, but following this convention often reduces sign errors and saves a step in the solution.
Using the Product-to-Difference identity
\displaystyle \cos A\sin B=\frac12\big[\sin(A+B)-\sin(A-B)\big]
Putting A=4x and B=3x, we get
\displaystyle \cos4x\sin3x=\frac12\big[\sin7x-\sin x\big]
Therefore,
\displaystyle I=\frac12\int\big[\sin7x-\sin x\big]\,dx
Using the properties of indefinite integrals, we get
\displaystyle I=\frac12\int\sin7x\,dx-\frac12\int\sin x\,dx
Using the standard formula,
\displaystyle \int\sin(ax+b)\,dx=-\frac{\cos(ax+b)}{a}+C
we get
\displaystyle I=\frac12\left(-\frac{\cos7x}{7}\right)-\frac12(-\cos x)+C
\displaystyle =-\frac{\cos7x}{14}+\frac{\cos x}{2}+C
Hence,
\boxed{\displaystyle I=-\frac{1}{14}\cos7x+\frac12\cos x+C}
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Likewise, this Integrals 7.3 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 Using Trigonometric Identities
3. Evaluate \int \cos2x\cos4x\cos6x\,dx.
Solution
Let, \displaystyle I=\int \cos2x\cos4x\cos6x\,dx
Rearranging the terms (for the same reason as in Question 2 above), we get
\displaystyle I=\int \cos4x\cos2x\cos6x\,dx
Using the Product-to-Sum identity in first two terms only
\displaystyle \cos A\cos B=\frac12\big[\cos(A+B)+\cos(A-B)\big]
Putting A=4x and B=2x, we get
\displaystyle \cos4x\cos2x=\frac12\big[\cos6x+\cos2x\big]
Therefore,
\displaystyle I=\frac12\int\big(\cos6x+\cos2x\big)\cos6x\,dx
\displaystyle I=\frac12\int\cos^26x\,dx+\frac12\int\cos2x\cos6x\,dx
Let \displaystyle I=\frac12\int\cos^26x\,dx+I_1 …(1)
where \displaystyle I_1=\frac12\int\cos2x\cos6x\,dx
Using the Power Reduction identity in \displaystyle \frac12\int\cos^26x\,dx
i.e. \displaystyle \cos^2A=\frac{1+\cos2A}{2}
with A=6x, we get
\displaystyle \cos^26x=\frac{1+\cos12x}{2}
Also, using the Product-to-Sum identity in I_1
i.e. \displaystyle \cos A\cos B=\frac12\big[\cos(A+B)+\cos(A-B)\big]
with A=6x and B=2x, we get
\displaystyle \cos6x\cos2x=\frac12\big[\cos8x+\cos4x\big]
Substituting these results in (1), we get
\displaystyle I=\frac12\int\frac12(1+\cos12x)\,dx+\frac12\int\frac12(\cos8x+\cos4x)\,dx
\displaystyle I=\frac14\int\big[1+\cos12x+\cos8x+\cos4x\big]\,dx
Using the standard formula,
\displaystyle \int\cos(ax+b)\,dx=\frac{\sin(ax+b)}{a}+C
we get
\displaystyle I=\frac14\left(x+\frac{\sin12x}{12}+\frac{\sin8x}{8}+\frac{\sin4x}{4}\right)+C
Hence,
\boxed{\displaystyle I=\frac14\big[\frac12\sin12x+\frac18\sin8x+\frac14\sin4x+x\big]+C}
Question 4: Integrals 7.3
4. Evaluate \int \sin^3(2x+1)\,dx.
Solution
Let, \displaystyle I=\int \sin^3(2x+1)\,dx
Rewriting the given integral as,
\displaystyle I=\int \sin(2x+1)\sin^2(2x+1)\,dx
Using the identity
\displaystyle \sin^2A=1-\cos^2A
we get
\displaystyle I=\int \sin(2x+1)\Big(1-\cos^2(2x+1)\Big)\,dx
\displaystyle I=\int \sin(2x+1)\,dx-\int \sin(2x+1)\cos^2(2x+1)\,dx
Let \displaystyle I=\int \sin(2x+1)\,dx-I_1 …(1)
where \displaystyle I_1=\int \sin(2x+1)\cos^2(2x+1)\,dx
Using the standard result,
\displaystyle \int\sin(ax+b)\,dx=-\frac{\cos(ax+b)}{a}+C
we get
\displaystyle \int \sin(2x+1)\,dx=-\frac12\cos(2x+1)
Now, to evaluate I_1, let
t=\cos(2x+1)
Then
\displaystyle \frac{dt}{dx}=-2\sin(2x+1)
or
\displaystyle \sin(2x+1)\,dx=-\frac12\,dt
Therefore,
\displaystyle I_1=-\frac12\int t^2\,dt
\displaystyle I_1=-\frac12\left(\frac{t^3}{3}\right)
or \displaystyle I_1=-\frac16\,t^3
Substituting t=\cos(2x+1), we get
\displaystyle I_1=-\frac16\cos^3(2x+1)
Putting this value of I_1 in equation (1), we get
\displaystyle I=-\frac12\cos(2x+1)-\left(-\frac16\cos^3(2x+1)\right)+C
\displaystyle I=-\frac12\cos(2x+1)+\frac16\cos^3(2x+1)+C
Hence,
\boxed{\displaystyle I=-\frac12\cos(2x+1)+\frac16\cos^3(2x+1)+C}
Alternative Method: Using the identity \displaystyle \sin^3A=\frac{3\sin A-\sin3A}{4}, we get
\displaystyle \sin^3(2x+1)=\frac{3\sin(2x+1)-\sin(6x+3)}{4}
Therefore,
\displaystyle \int\sin^3(2x+1)\,dx=\frac14\int\Big(3\sin(2x+1)-\sin(6x+3)\Big)\,dx
Evaluating the integrals, we obtain
\displaystyle \int\sin^3(2x+1)\,dx=-\frac38\cos(2x+1)+\frac1{24}\cos(6x+3)+C
This answer is equivalent to the NCERT answer and differs only in form, which can be easily checked by using cos 3A identity.
Question 5: Integrals 7.3
5. Evaluate \int \sin^3x\cos^3x\,dx.
Solution
Let, \displaystyle I=\int \sin^3x\cos^3x\,dx
Write \sin^3x as \sin x\sin^2x and use the identity
\displaystyle \sin^2x=1-\cos^2x
We get
\displaystyle I=\int \sin x(1-\cos^2x)\cos^3x\,dx
\displaystyle I=\int \sin x(\cos^3x-\cos^5x)\,dx
Using the properties of indefinite integrals,
\displaystyle I=\int \sin x\cos^3x\,dx-\int \sin x\cos^5x\,dx
Let \displaystyle I=I_1-I_2 …(1)
where
\displaystyle I_1=\int \sin x\cos^3x\,dx
and
\displaystyle I_2=\int \sin x\cos^5x\,dx
To evaluate I_1, let
t=\cos x
Then
\displaystyle dt=-\sin x\,dx
Therefore,
\displaystyle I_1=\int \sin x\cos^3x\,dx=-\int t^3\,dt
\displaystyle I_1=-\frac{t^4}{4}
i.e. \displaystyle I_1=-\frac{\cos^4x}{4} …(2)
Similarly, for I_2, let
t=\cos x
Then
\displaystyle dt=-\sin x\,dx
Therefore,
\displaystyle I_2=\int \sin x\cos^5x\,dx=-\int t^5\,dt
\displaystyle I_2=-\frac{t^6}{6}
i.e. \displaystyle I_2=-\frac{\cos^6x}{6} …(3)
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=-\frac{\cos^4x}{4}-\left(-\frac{\cos^6x}{6}\right)+C
\displaystyle I=-\frac{\cos^4x}{4}+\frac{\cos^6x}{6}+C
Hence,
\boxed{\displaystyle I=\frac16\cos^6x-\frac14\cos^4x+C}
Important: The real challenge in this exercise is usually not the integration itself, but identifying the correct trigonometric identity to simplify the integrand. Once the expression is converted into a simpler form, the required integral can usually be evaluated using standard formulas.
Question 6: Integrals Exercise 7.3
6. Evaluate \int \sin x\sin2x\sin3x\,dx.
Solution
Let, \displaystyle I=\int \sin x\sin2x\sin3x\,dx
Rearranging the terms, we get
\displaystyle I=\int \sin3x\sin x\sin2x\,dx
Using the Product-to-Sum identity in first two terms only
\displaystyle \sin A\sin B=\frac12\big[\cos(A-B)-\cos(A+B)\big]
Putting A=3x and B=x, we get
\displaystyle \sin3x\sin x=\frac12\big[\cos2x-\cos4x\big]
Therefore,
\displaystyle I=\frac12\int\big(\cos2x-\cos4x\big)\sin2x\,dx
or \displaystyle I=\frac12\int\sin2x\cos2x\,dx-\frac12\int\sin2x\cos4x\,dx …(1)
Using the Double-Angle identity in first integral
\displaystyle \sin2A=2\sin A\cos A
with A=2x, we get
\displaystyle \sin2x\cos2x=\frac12\sin4x
Also, using the Product-to-Sum identity in the second integral
\displaystyle \sin A\cos B=\frac12\big[\sin(A+B)+\sin(A-B)\big]
Putting A=4x and B=2x, we get
\displaystyle \sin2x\cos4x=\frac12\big[\sin6x-\sin2x\big]
Substituting these results in (1), we get
\displaystyle I=\frac12\int\frac12\sin4x\,dx-\frac12\int\frac12\big(\sin6x-\sin2x\big)\,dx
\displaystyle I=\frac14\int\big[\sin4x-\sin6x+\sin2x\big]\,dx
Using the standard formula,
\displaystyle \int\sin(ax+b)\,dx=-\frac{\cos(ax+b)}{a}+C
we get
\displaystyle I=\frac14\left(-\frac{\cos4x}{4}+\frac{\cos6x}{6}-\frac{\cos2x}{2}\right)+C
Hence,
\boxed{\displaystyle I=\frac14\left(\frac16\cos6x-\frac14\cos4x-\frac12\cos2x\right)+C}
Question 7: Exercise 7.3
7. Evaluate \int \sin4x\sin8x\,dx.
Solution
Let, \displaystyle I=\int \sin4x\sin8x\,dx
Rewriting the give integral as
\displaystyle I=\int \sin8x\sin4x\,dx
Using the Product-to-Sum/Difference identity
\displaystyle \sin A\sin B=\frac12\big[\cos(A-B)-\cos(A+B)\big]
Putting A=8x and B=4x, we get
\displaystyle \sin8x\sin4x=\frac12\big[\cos4x-\cos12x\big]
Therefore,
\displaystyle I=\frac12\int\big(\cos4x-\cos12x\big)\,dx
Using the properties of indefinite integrals, we get
\displaystyle I=\frac12\int\cos4x\,dx-\frac12\int\cos12x\,dx
Using the standard formula,
\displaystyle \int\cos(ax+b)\,dx=\frac{\sin(ax+b)}{a}+C
we get
\displaystyle I=\frac12\left(\frac{\sin4x}{4}\right)-\frac12\left(\frac{\sin12x}{12}\right)+C
\displaystyle I=\frac12\left(\frac{\sin4x}{4}-\frac{\sin12x}{12}\right)+C
Hence,
\boxed{\displaystyle I=\frac12\left(\frac14\sin4x-\frac{1}{12}\sin12x\right)+C}
Question 8: Integration Using Trigonometric Identities
8. Evaluate \int\frac{1-\cos x}{1+\cos x}\,dx.
Solution
Let, \displaystyle I=\int\frac{1-\cos x}{1+\cos x}\,dx
Using the identity
\displaystyle \frac{1-\cos x}{1+\cos x}=\frac{\cancel{2}\sin^2\frac{x}{2}}{\cancel{2}\cos^2\frac{x}{2}}=\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}=\tan^2\frac{x}{2}
we get
\displaystyle I=\int\tan^2\frac{x}{2}\,dx
Using the identity
\displaystyle \tan^2\theta=\sec^2\theta-1
with \theta=\frac{x}{2}, we get
\displaystyle I=\int\left(\sec^2\frac{x}{2}-1\right)\,dx
Using the properties of indefinite integrals,
\displaystyle I=\int\sec^2\frac{x}{2}\,dx-\int1\,dx
Using the standard formula
\displaystyle \int\sec^2(ax+b)\,dx=\frac{\tan(ax+b)}{a}+C
we get
\displaystyle I=2\tan\frac{x}{2}-x+C
Hence,
\boxed{\displaystyle I=2\tan\frac{x}{2}-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.
Question 9: Integrals 7.3
9. Evaluate \int\frac{\cos x}{1+\cos x}\,dx.
Solution
Let, \displaystyle I=\int\frac{\cos x}{1+\cos x}\,dx
Adding and subtracting 1 in the numerator, we get
\displaystyle I=\int\frac{1+\cos x-1}{1+\cos x}\,dx
Splitting the numerator, we get
\displaystyle I=\int\left(1-\frac{1}{1+\cos x}\right)\,dx
Using the identity
\displaystyle 1+\cos x=2\cos^2\frac{x}{2}
we get
\displaystyle I=\int\left(1-\frac{1}{2\cos^2(x/2)}\right)\,dx
\displaystyle I=\int\left(1-\frac12\sec^2\frac{x}{2}\right)\,dx
Using the properties of indefinite integrals,
\displaystyle I=\int1\,dx-\frac12\int\sec^2\frac{x}{2}\,dx
Using the standard formula
\displaystyle \int\sec^2(ax+b)\,dx=\frac{\tan(ax+b)}{a}+C
we get
\displaystyle I=x-\frac12\left(\frac{\tan\frac{x}{2}}{\frac12}\right)+C
\displaystyle I=x-\tan\frac{x}{2}+C
Hence,
\boxed{\displaystyle I=x-\tan\frac{x}{2}+C}
Question 10: Integrals 7.3
10. Evaluate \int \sin^4x\,dx.
Solution
Let, \displaystyle I=\int \sin^4x\,dx
Writing \sin^4x as (\sin^2x)^2, we get
\displaystyle I=\int (\sin^2x)^2\,dx
Using the Power Reduction identity
\displaystyle \sin^2x=\frac{1-\cos2x}{2}
we get
\displaystyle I=\int \left(\frac{1-\cos2x}{2}\right)^2dx
\displaystyle I=\frac14\int(1-\cos2x)^2\,dx
Expanding the square,
\displaystyle I=\frac14\int\big(1-2\cos2x+\cos^22x\big)\,dx
\displaystyle I=\frac14\left(\int1\,dx-2\int\cos2x\,dx+\int\cos^22x\,dx\right)
Let \displaystyle I=\frac14\left(\int1\,dx-2\int\cos2x\,dx+I_1\right)\qquad ...(1)
where \displaystyle I_1=\int\cos^22x\,dx
Integrating the first two integrals in eq. (1), using standard formulas, we get
\displaystyle I=\frac14\left(x-2\left(\frac{\sin2x}{2}\right)+I_1\right)+C
or \displaystyle I=\frac14\left(x-\sin2x+I_1\right)+C\qquad ...(2)
Now, to evaluate I_1, use the Power Reduction identity
\displaystyle \cos^2A=\frac{1+\cos2A}{2}
with A=2x. Thus,
\displaystyle I_1=\int\frac{1+\cos4x}{2}\,dx
\displaystyle I_1=\frac12\int1\,dx+\frac12\int\cos4x\,dx
Using the standard formula
\displaystyle \int\cos(ax+b)\,dx=\frac{\sin(ax+b)}{a}+C
we get
\displaystyle I_1=\frac{x}{2}+\frac{\sin4x}{8} …(3)
Putting the value of I_1 from (3) into (2), we get
\displaystyle I=\frac14\left(x-\sin2x+\frac{x}{2}+\frac{\sin4x}{8}\right)+C
\displaystyle I=\frac{3x}{8}-\frac{\sin2x}{4}+\frac{\sin4x}{32}+C
Hence,
\boxed{\displaystyle I=\frac{3x}{8}-\frac14\sin2x+\frac{1}{32}\sin4x+C}
If you’ve reached this point, you’ve already practiced several important trigonometric identities. The good news is that most remaining questions follow the same pattern: first simplify the integrand using an appropriate identity, then apply the standard integration formulas. With a little practice, identifying the right identity becomes much easier.
Now, let’s move on to the next question.
Question 11: Integration of Trigonometric Functions
11. Evaluate \int \cos^4 2x\,dx.
Solution
Let, \displaystyle I=\int \cos^4 2x\,dx
Writing \cos^4 2x as (\cos^2 2x)^2, we get
\displaystyle I=\int (\cos^2 2x)^2\,dx
Using the Power Reduction identity
\displaystyle \cos^2A=\frac{1+\cos2A}{2}
with A=2x, we get
\displaystyle I=\int \left(\frac{1+\cos4x}{2}\right)^2dx
\displaystyle I=\frac14\int(1+\cos4x)^2\,dx
Expanding the square,
\displaystyle I=\frac14\int\big(1+2\cos4x+\cos^24x\big)\,dx
\displaystyle I=\frac14\left(\int1\,dx+2\int\cos4x\,dx+\int\cos^24x\,dx\right)
Let \displaystyle I=\frac14\left(\int1\,dx+2\int\cos4x\,dx+I_1\right)\qquad ...(1)
where \displaystyle I_1=\int\cos^24x\,dx
Integrating the first two integrals in eq. (1), using standard formulas, we get
\displaystyle I=\frac14\left(x+2\left(\frac{\sin4x}{4}\right)+I_1\right)+C
\displaystyle I=\frac14\left(x+\frac12\sin4x+I_1\right)+C\qquad ...(2)
Now, to evaluate I_1, use the Power Reduction identity
\displaystyle \cos^2A=\frac{1+\cos2A}{2}
with A=4x. Thus,
\displaystyle I_1=\int\frac{1+\cos8x}{2}\,dx
\displaystyle I_1=\frac12\int1\,dx+\frac12\int\cos8x\,dx
Using the standard formula
\displaystyle \int\cos(ax+b)\,dx=\frac{\sin(ax+b)}{a}+C
we get
\displaystyle I_1=\frac{x}{2}+\frac{\sin8x}{16} …(3)
Putting the value of I_1 from (3) into (2), we get
\displaystyle I=\frac14\left(x+\frac{\sin4x}{2}+\frac{x}{2}+\frac{\sin8x}{16}\right)+C
\displaystyle I=\frac{3x}{8}+\frac{\sin4x}{8}+\frac{\sin8x}{64}+C
Hence,
\boxed{\displaystyle I=\frac{3x}{8}+\frac18\sin4x+\frac{1}{64}\sin8x+C}
Question 12: Integrals 7.3
12. Evaluate \int\frac{\sin^2x}{1+\cos x}\,dx.
Solution
Let, \displaystyle I=\int\frac{\sin^2x}{1+\cos x}\,dx
Using the identity
\displaystyle \sin^2x=(1-\cos x)(1+\cos x)
we get
\displaystyle I=\int\frac{(1-\cos x)(1+\cos x)}{1+\cos x}\,dx
\displaystyle I=\int(1-\cos x)\,dx
Using the properties of indefinite integrals, we get
\displaystyle I=\int1\,dx-\int\cos x\,dx
i.e. \displaystyle I=x-\sin x+C
Hence,
\boxed{\displaystyle I=x-\sin x+C}
Question 13: Integrals 7.3
13. Evaluate \int\frac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha}\,dx.
Solution
Let, \displaystyle I=\int\frac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha}\,dx
Using the identity
\displaystyle \cos2A=2\cos^2A-1
we get
\displaystyle I=\int\frac{(2\cos^2x-1)-(2\cos^2\alpha-1)}{\cos x-\cos\alpha}\,dx
\displaystyle I=2\int\frac{\cos^2x-\cos^2\alpha}{\cos x-\cos\alpha}\,dx
Using the identity
\displaystyle a^2-b^2=(a-b)(a+b)
with a=\cos x and b=\cos\alpha, we get
\displaystyle I=2\int\frac{(\cos x-\cos\alpha)(\cos x+\cos\alpha)}{\cos x-\cos\alpha}\,dx
\displaystyle I=2\int(\cos x+\cos\alpha)\,dx
Using the properties of indefinite integrals,
\displaystyle I=2\int\cos x\,dx+2\cos\alpha\int dx
Note that \cos\alpha is a constant here.
So, we get
\displaystyle I=2\sin x+2x\cos\alpha+C
Hence,
\boxed{\displaystyle I=2(\sin x+x\cos\alpha)+C}
Question 14: Integrals Exercise 7.3
14. Evaluate \int\frac{\cos x-\sin x}{1+\sin2x}\,dx.
Solution
Let, \displaystyle I=\int\frac{\cos x-\sin x}{1+\sin2x}\,dx
Using the identity
\displaystyle \sin2x=2\sin x\cos x
we get
\displaystyle I=\int\frac{\cos x-\sin x}{1+2\sin x\cos x}\,dx
Since
\displaystyle (\sin x+\cos x)^2=\sin^2x+\cos^2x+2\sin x\cos x=1+2\sin x\cos x
therefore,
\displaystyle I=\int\frac{\cos x-\sin x}{(\sin x+\cos x)^2}\,dx
Let
t=\sin x+\cos x
Then
\displaystyle \frac{dt}{dx}=\cos x-\sin x
or
\displaystyle dt=(\cos x-\sin x)\,dx
Therefore, \displaystyle I=\int\frac{dt}{t^2}=\int t^{-2}\,dt
\displaystyle I=\frac{t^{-1}}{-1}+C
\displaystyle I=-\frac1t+C
Substituting t=\sin x+\cos x, we get
\displaystyle I=-\frac1{\sin x+\cos x}+C
Hence,
\boxed{\displaystyle I=-\frac1{\sin x+\cos x}+C}
Interesting Fact: Many trigonometric identities that you learn in Class 11, such as Product-to-Sum, Sum-to-Product and Double-Angle formulas, become powerful tools in integration. Concepts from different chapters often come together while solving advanced calculus problems.
Question 15: Indefinite Integrals 7.3
15. Evaluate \int \tan^3 2x\sec2x\,dx.
Solution
Let, \displaystyle I=\int \tan^3 2x\sec2x\,dx
Writing \tan^3 2x as \tan2x\cdot\tan^22x, we get
\displaystyle I=\int \tan2x\cdot\tan^22x\cdot\sec2x\,dx
💡We do this so that the given integral can be converted into a recognisable form, making it easier to apply a suitable identity and evaluate the integral.
Now, using the identity
\displaystyle \tan^2A=\sec^2A-1
with A=2x, we get
\displaystyle I=\int \tan2x(\sec^22x-1)\sec2x\,dx
\displaystyle I=\int \tan2x\sec^32x\,dx-\int \tan2x\sec2x\,dx
or \displaystyle I=I_1-I_2 …(1)
where \displaystyle I_1=\int \tan2x\sec^32x\,dx
and
\displaystyle I_2=\int \tan2x\sec2x\,dx
Now, to evaluate I_1, let
t=\sec2x
Then
\displaystyle dt=2\sec2x\tan2x\,dx
or
\displaystyle \sec2x\tan2x\,dx=\frac{dt}{2}
Therefore, \displaystyle I_1=\int \tan2x\sec^32x\,dx=\frac12\int t^2\,dt
\displaystyle I_1=\frac12\cdot\frac{t^3}{3}
\displaystyle I_1=\frac{\sec^32x}{6}\qquad ...(2)
Similarly, for I_2, let
t=\sec2x
Then
\displaystyle \sec2x\tan2x\,dx=\frac{dt}{2}
Therefore, \displaystyle I_2=\int \tan2x\sec2x\,dx=\frac12\int dt
\displaystyle I_2=\frac{t}{2}
\displaystyle I_2=\frac{\sec2x}{2}\qquad ...(3)
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=\frac{\sec^32x}{6}-\frac{\sec2x}{2}+C
Hence,
\boxed{\displaystyle I=\frac16\sec^32x-\frac12\sec2x+C}
Interesting Fact: Many trigonometric integrals that appear complicated at first can be evaluated by applying a suitable identity. In fact, a large part of integration involves recognizing patterns and transforming the integrand into a simpler form before applying standard integration formulas.
Question 16: Integrals 7.3
16. Evaluate \int \tan^4x\,dx.
Solution
Let, \displaystyle I=\int \tan^4x\,dx
Writing \tan^4x as (\tan^2x)^2, we get
\displaystyle I=\int (\tan^2x)^2\,dx
Using the identity
\displaystyle \tan^2x=\sec^2x-1
we get
\displaystyle I=\int (\sec^2x-1)^2\,dx
Expanding the square,
\displaystyle I=\int (\sec^4x-2\sec^2x+1)\,dx
Using the properties of indefinite integrals, we get
\displaystyle I=\int\sec^4x\,dx-2\int\sec^2x\,dx+\int1\,dx
\displaystyle \because \int\sec^2x\,dx=\tan x and \displaystyle \int1\,dx=x
We get, \displaystyle I=\int\sec^4x\,dx-\tan x+x+C
Further, let \displaystyle I=I_1-\tan x+x+C\qquad ...(1)
where \displaystyle I_1=\int\sec^4x\,dx
To evaluate I_1, write
\displaystyle I_1=\int\sec^2x\cdot\sec^2x\,dx
Using the identity
\displaystyle \sec^2x=1+\tan^2x
we get
\displaystyle I_1=\int(1+\tan^2x)\sec^2x\,dx
Let
t=\tan x
Then
\displaystyle dt=\sec^2x\,dx
Therefore, \displaystyle I_1=\int(1+t^2)\,dt
\displaystyle I_1=t+\frac{t^3}{3}
i.e. \displaystyle I_1=\tan x+\frac{\tan^3x}{3}\qquad ...(2)
Putting the value of I_1 from (2) into (1), we get
\displaystyle I=\left(\tan x+\frac{\tan^3x}{3}\right)-2\tan x+x+C
\displaystyle I=\frac{\tan^3x}{3}-\tan x+x+C
Hence,
\boxed{\displaystyle I=\frac13\tan^3x-\tan x+x+C}
Question 17: Integration 7.3
17. Evaluate \int\frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\,dx.
Solution
Let, \displaystyle I=\int\frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\,dx
Separating the terms, we get
\displaystyle I=\int\left(\frac{\sin^3x}{\sin^2x\cos^2x}+\frac{\cos^3x}{\sin^2x\cos^2x}\right)dx
Cancelling the common factors, we get
\displaystyle I=\int\left(\frac{\sin x}{\cos^2x}+\frac{\cos x}{\sin^2x}\right)dx
or \displaystyle I=\int\frac{\sin x}{\cos^2x}\,dx+\int\frac{\cos x}{\sin^2x}\,dx
Let, \displaystyle I=I_1+I_2 \qquad ...(1)
where
\displaystyle I_1=\int\frac{\sin x}{\cos^2x}\,dx
and
\displaystyle I_2=\int\frac{\cos x}{\sin^2x}\,dx
Now, to evaluate I_1, let
t=\cos x
Then
\displaystyle dt=-\sin x\,dx
Therefore, \displaystyle I_1=\int\frac{\sin x}{\cos^2x}\,dx=-\int\frac{dt}{t^2}
\displaystyle I_1=-\int t^{-2}\,dt=-\frac{t^{-1}}{-1}
\displaystyle I_1=\frac1t=\frac{1}{\cos x}
So, \displaystyle I_1=\sec x \qquad ...(2)
To evaluate I_2, let
t=\sin x
Then
\displaystyle dt=\cos x\,dx
Therefore, \displaystyle I_2=\int\frac{\cos x}{\sin^2x}\,dx=\int\frac{dt}{t^2}
\displaystyle I_2=\int t^{-2}\,dt=\frac{t^{-1}}{-1}
\displaystyle I_2=-\frac1t=-\frac{1}{\sin x}
i.e. \displaystyle I_2=-\cosec x \qquad ...(3)
Putting the values of I_1 and I_2 from (2) and (3) into (1), we get
\displaystyle I=\sec x-\cosec x+C
Hence,
\boxed{\displaystyle I=\sec x-\cosec x+C}
Question 18: Integration Using Trigonometric Identities
18. Evaluate \int\frac{\cos2x+2\sin^2x}{\cos^2x}\,dx.
Solution
Let, \displaystyle I=\int\frac{\cos2x+2\sin^2x}{\cos^2x}\,dx
Using the identity
\displaystyle \cos2x=\cos^2x-\sin^2x
we get
\displaystyle I=\int\frac{\cos^2x-\sin^2x+2\sin^2x}{\cos^2x}\,dx
\displaystyle I=\int\frac{\cos^2x+\sin^2x}{\cos^2x}\,dx
Using the identity
\displaystyle \sin^2x+\cos^2x=1
we get
\displaystyle I=\int\frac{1}{\cos^2x}\,dx
\displaystyle I=\int\sec^2x\,dx
Using the standard formula
\displaystyle \int\sec^2x\,dx=\tan x+C
we get
\displaystyle I=\tan x+C
Hence,
\boxed{\displaystyle I=\tan x+C}
Question 19: Integrals Exercise 7.3
19. Evaluate \int\frac{1}{\sin x\cos^3x}\,dx.
Solution
Let, \displaystyle I=\int\frac{1}{\sin x\cos^3x}\,dx
Writing the integrand suitably, we get
\displaystyle I=\int\frac{1}{\sin x\cos x\cos^2x}\,dx
or \displaystyle I=\int\frac{\sec^2x}{\sin x\cos x}\,dx
Using the identity
\displaystyle \sin2x=2\sin x\cos x
we get
\displaystyle I=\int\frac{2\sec^2x}{\sin2x}\,dx
Now, using the identity
\displaystyle \sin2x=\frac{2\tan x}{1+\tan^2x}
we get
\displaystyle I=\int\frac{2\sec^2x}{\frac{2\tan x}{1+\tan^2x}}\,dx
\displaystyle I=\int\frac{1+\tan^2x}{\tan x}\sec^2x\,dx
We have now expressed the integrand in terms of \tan x. Also, the derivative of \tan x is \sec^2x, which is already present in the integral.
Let
t=\tan x
Then
\displaystyle dt=\sec^2x\,dx
Therefore, \displaystyle I=\int\frac{1+t^2}{t}\,dt
\displaystyle I=\int\left(t+\frac1t\right)dt
\displaystyle I=\frac{t^2}{2}+\log|t|+C
Substituting t=\tan x, we get
\displaystyle I=\frac{\tan^2x}{2}+\log|\tan x|+C
Hence,
\boxed{\displaystyle I=\frac12\tan^2x+\log|\tan x|+C}
Question 20: Trigonometric Integrals
20. Evaluate \int\frac{\cos2x}{(\cos x+\sin x)^2}\,dx.
Solution
Let, \displaystyle I=\int\frac{\cos2x}{(\cos x+\sin x)^2}\,dx
Using the identity
\displaystyle \cos2x=\cos^2x-\sin^2x
and the factorization
\displaystyle a^2-b^2=(a+b)(a-b)
we get
\displaystyle \cos2x=(\cos x+\sin x)(\cos x-\sin x)
Therefore,
\displaystyle I=\int\frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^2}\,dx
\displaystyle I=\int\frac{\cos x-\sin x}{\cos x+\sin x}\,dx
Let
t=\cos x+\sin x
Then
\displaystyle dt=(\cos x-\sin x)\,dx
Therefore,
\displaystyle I=\int\frac{dt}{t}
\displaystyle I=\log|t|+C
Substituting t=\cos x+\sin x, we get
\displaystyle I=\log|\cos x+\sin x|+C
Hence,
\boxed{\displaystyle I=\log|\cos x+\sin x|+C}
Question 21: Basic Integrals 7.3
21. Evaluate \int \sin^{-1}(\cos x)\,dx.
Solution
Since \cos x=\sin\left(\frac{\pi}{2}-x\right), therefore \sin^{-1}(\cos x)=\frac{\pi}{2}-x.
So, the given integral reduces to
\displaystyle I=\int\left(\frac{\pi}{2}-x\right)\,dx
Using the properties of indefinite integrals,
\displaystyle I=\frac{\pi}{2}\int dx-\int x\,dx
Using the standard formula
\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C
we get
\displaystyle I=\frac{\pi x}{2}-\frac{x^2}{2}+C
Hence,
\boxed{\displaystyle I=\frac{\pi x}{2}-\frac{x^2}{2}+C}
Question 22: Important Integrals
22. Evaluate \int\frac{dx}{\cos(x-a)\cos(x-b)}.
Solution
Let, \displaystyle I=\int\frac{dx}{\cos(x-a)\cos(x-b)}
Multiplying and dividing by \sin(a-b), we get
\displaystyle I=\frac1{\sin(a-b)}\int\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}\,dx
Now,
\displaystyle a-b=(x-b)-(x-a)
Therefore,
\displaystyle \sin(a-b)=\sin\big((x-b)-(x-a)\big)
Using the identity
\displaystyle \sin(A-B)=\sin A\cos B-\cos A\sin B
with A=x-b and B=x-a, we get
\displaystyle \sin(a-b)=\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)
Substituting in the integral, we get
I=\frac1{\sin(a-b)}\int\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}\,dx
Splitting the numerator and canceling the common factors, we get
\displaystyle I=\frac1{\sin(a-b)}\int\Big[\tan(x-b)-\tan(x-a)\Big]\,dx
Using the standard formula
\displaystyle \int\tan(ax+b)\,dx=-\frac1a\log|\cos(ax+b)|+C
we get
\displaystyle I=\frac1{\sin(a-b)}\Big[-\log|\cos(x-b)|+\log|\cos(x-a)|\Big]+C
By using the property of logarithms,
\displaystyle \log m-\log n=\log\left(\frac{m}{n}\right), we get
\displaystyle I=\frac1{\sin(a-b)}\log\left|\frac{\cos(x-a)}{\cos(x-b)}\right|+C
Hence,
\boxed{\displaystyle I=\frac1{\sin(a-b)}\log\left|\frac{\cos(x-a)}{\cos(x-b)}\right|+C}
Useful Observation: In integrals involving products such as \cos(x-a)\cos(x-b) or \sin(x-a)\sin(x-b) in the denominator, it is often helpful to multiply and divide by \sin(a-b). Then write \sin(a-b)=\sin\big((x-b)-(x-a)\big) and apply the identity \sin(A-B). Similarly, when the denominator contains unlike functions such as \sin(x-a)\cos(x-b) or \cos(x-a)\sin(x-b), multiplying and dividing by \cos(a-b) and then using \cos(A-B) is often the most effective approach.
Question 23: Integrals 7.3 – MCQ
23. Choose the correct answer: \displaystyle \int\frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\,dx is equal to
- (A) \tan x+\cot x+C
- (B) \tan x+\cosec x+C
- (C) -\tan x+\cot x+C
- (D) \tan x+\sec x+C
Solution
Let,
\displaystyle I=\int\frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\,dx
Separating the terms, we get
\displaystyle I=\int\left(\frac{\sin^2x}{\sin^2x\cos^2x}-\frac{\cos^2x}{\sin^2x\cos^2x}\right)dx
\displaystyle I=\int\left(\sec^2x-\csc^2x\right)dx
Using the properties of indefinite integrals,
\displaystyle I=\int\sec^2x\,dx-\int\cosec^2x\,dx
Using the standard formulas
\displaystyle \int\sec^2x\,dx=\tan x
and
\displaystyle \int\cosec^2x\,dx=-\cot x
we get
\displaystyle I=\tan x-(-\cot x)+C
\displaystyle I=\tan x+\cot x+C
✅️ Hence, the correct answer is (A)
\boxed{\tan x+\cot x+C}
Question 24: Integrals 7.3 – MCQ
24. Choose the correct answer: \displaystyle \int\frac{e^x(1+x)}{\cos^2(e^xx)}\,dx equals
- (A) -\cot(ex^x)+C
- (B) \tan(xe^x)+C
- (C) \tan(e^x)+C
- (D) \cot(e^x)+C
Solution
Let,
\displaystyle I=\int\frac{e^x(1+x)}{\cos^2(xe^x)}\,dx
Since \displaystyle \frac{1}{\cos^2(xe^x)}=\sec^2(xe^x), we get
\displaystyle I=\int e^x(1+x)\sec^2(xe^x)\,dx
Let
t=xe^x
Then, using the product rule,
\displaystyle \frac{dt}{dx}=e^x+xe^x=e^x(1+x)
or
\displaystyle dt=e^x(1+x)\,dx
Therefore,
\displaystyle I=\int\sec^2 t\,dt
Using the standard formula
\displaystyle \int\sec^2 t\,dt=\tan t+C
we get
\displaystyle I=\tan t+C
Substituting t=xe^x, we get
\displaystyle I=\tan(xe^x)+C
✅️ Hence, the correct answer is (B)
\boxed{\tan(xe^x)+C}
Common Mistakes to Avoid
- Forgetting to simplify using trigonometric identities first: Many integrals in this exercise cannot be evaluated directly. Always check whether identities such as \sin^2A=\frac{1-\cos2A}{2}, \cos^2A=\frac{1+\cos2A}{2} or Product-to-Sum formulas can simplify the integrand.
- Using the wrong Product-to-Sum identity: Be careful while applying identities such as \sin A\cos B, \cos A\cos B and \sin A\sin B. A small sign error can completely change the answer.
- Ignoring the order while applying Sum-to-Product formulas: To avoid sign mistakes, it is often helpful to assume A>B (or C>D) and write identities accordingly.
- Forgetting Power Reduction identities: Integrals involving \sin^2x, \cos^2x, \sin^4x or \cos^4x are usually simplified using Power Reduction formulas before integration.
- Missing obvious simplifications: Before applying advanced identities, check whether the numerator and denominator can be simplified directly using identities such as \sin^2x+\cos^2x=1 or factorization formulas.
- Forgetting the coefficient while integrating trigonometric functions: For expressions such as \sin(ax+b) or \cos(ax+b), remember to divide by the coefficient of x. For example, \displaystyle \int\sin(ax+b)\,dx=-\frac{\cos(ax+b)}{a}+C.
- Making mistakes in substitution: When using substitution, write t and dt clearly and substitute every occurrence of the variable before integrating.
- Not separating complicated integrals: If an integral contains two or more terms, split it into simpler integrals using the properties of indefinite integrals before proceeding further.
- Forgetting the constant of integration: Every indefinite integral represents a family of functions and must include the arbitrary constant C.
- Not verifying the final answer: If time permits, differentiate the result. If differentiation gives back the original integrand, the integration is correct.
Continue Learning
After completing Exercise 7.3 of Integrals, you should now be comfortable with evaluating integrals by using a variety of trigonometric identities. Many integrals that appear complicated at first become straightforward after applying the appropriate identity and simplifying the integrand before integration.
To strengthen your understanding further, make sure that you revise:
- The Product-to-Sum identities for \sin A\sin B, \cos A\cos B and \sin A\cos B
- The Sum-to-Product identities for \sin C\pm\sin D and \cos C\pm\cos D
- Double-angle identities involving \sin2A, \cos2A and \tan2A
- Power Reduction formulas such as \displaystyle \sin^2A=\frac{1-\cos2A}{2} and \displaystyle \cos^2A=\frac{1+\cos2A}{2}
- Expressing higher powers like \sin^3A, \cos^3A, \sin^4A and \cos^4A in a suitable form before integration
- Recognising when a trigonometric identity simplifies the integrand into a standard integral
- Using substitution after simplification whenever the integrand contains a function along with its derivative
- Splitting complicated integrals into simpler integrals using the properties of indefinite integrals
- The standard formulas for \int\sin(ax+b)\,dx, \int\cos(ax+b)\,dx, \int\sec^2x\,dx and \int\cosec^2x\,dx
- Checking answers by differentiating the final result whenever possible
Explore More
Great job on completing Integrals 7.3! In this exercise, we learned how to simplify integrals using trigonometric identities such as double-angle, triple-angle, product-to-sum, sum-to-product, and power-reduction formulas. Identifying the correct identity before integrating is often the key to solving these questions efficiently.
If you found these solutions helpful, consider subscribing to my YouTube Channel, @MathsBetter, where many NCERT questions are explained step-by-step in video format. You can now continue with the next exercise to explore more techniques of integration.
All the best and keep learning 👍
