Integration using Trigonometric Identities Maths12: Imp Questions 2024

When dealing with Integration using Trigonometric Identities , direct integration can be difficult or even impossible without first transforming the expression., Trigonometric identities are incredibly useful in solving integration problems because they help simplify complex expressions into more manageable forms.

For example, identities like

$\sin^2x + \cos^2x = 1$

or,

$\sin 2x = 2 \sin x \cos x$

$sin 2 x = 2 sin x cos x$ allows us to rewrite terms in a way that makes the integral easier to solve. These identities also help in converting trigonometric functions into algebraic forms, enabling techniques like substitution to be applied more effectively.

Using trigonometric identities often reduces the complexity of the integral and makes it possible to integrate functions that might otherwise be challenging or time-consuming to solve directly.

Here are some commonly used trigonometric identities that are helpful in solving integration problems:

**Integration using Trigonometric Identities**

Pythagorean Identities:
$\sin^2 x + \cos^2 x = 1$
$1 + \tan^2 x = \sec^2 x$

$1 + \cot^2 x = \csc^2 x$
Double Angle Identities:
$\sin 2x = 2 \sin x \cos x$
$\cos 2x = \cos^2 x – \sin^2 x$

$\cos 2x = 2 \cos^2 x – 1$

$\cos 2x = 1 – 2 \sin^2 x$
Half Angle Identities:
$\sin^2 x = \frac{1 – \cos 2x}{2}$
$\cos^2 x = \frac{1 + \cos 2x}{2}$
Product-to-Sum Identities:
$\sin x \sin y = \frac{1}{2} [\cos (x – y) – \cos (x + y)]$
$\cos x \cos y = \frac{1}{2} [\cos (x – y) + \cos (x + y)]$

$\sin x \cos y = \frac{1}{2} [\sin (x + y) + \sin (x – y)]$
Basic Trigonometric Function Derivatives/Integrals:
$\frac{d}{dx} (\sin x) = \cos x, \quad \int \cos x \, dx = \sin x + C$
$\frac{d}{dx} (\cos x) = -\sin x, \quad \int \sin x \, dx = -\cos x + C$

$\int \sec^2 x \, dx = \tan x + C$

$\int \csc^2 x \, dx = -\cot x + C$

Integration using Trigonometric Identities
Sum and Difference Identities:
$\sin (x \pm y) = \sin x \cos y \pm \cos x \sin y$
$\cos (x \pm y) = \cos x \cos y \mp \sin x \sin y$

$\tan (x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y}$

7. Co-function Identities:

$\sin \left( \frac{\pi}{2} – x \right) = \cos x$

$\cos \left( \frac{\pi}{2} – x \right) = \sin x$

$\tan \left( \frac{\pi}{2} – x \right) = \cot x$

$\cot \left( \frac{\pi}{2} – x \right) = \tan x$

$\sec \left( \frac{\pi}{2} – x \right) = \csc x$

$\csc \left( \frac{\pi}{2} – x \right) = \sec x$

8. Inverse Trigonometric Identities:

$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$

$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$

$\sec^{- 1} x + \csc^{- 1} x = \frac{π}{2}$

9. Triple Angle Formula for Sine:

$\sin 3 x = 3 \sin x - 4 \sin^{}$

10. Triple Angle Formula for Cosine:

$\cos 3x = 4 \cos^3 x – 3 \cos x$

11. Triple Angle Formula for Tangent:

$\tan 3x = \frac{3 \tan x – \tan^3 x}{1 – 3 \tan^2 x}$

Reciprocal Identities:

These identities relate the basic trigonometric functions to their reciprocals.

$\csc x = \frac{1}{\sin x}, \quad \sec x = \frac{1}{\cos x}, \quad \cot x = \frac{1}{\tan x}$

13. Integral of Trigonometric Functions:

These are directly useful in solving integrals:

$\int \sin ax \, dx = -\frac{1}{a} \cos ax + C$

$\int \cos ax \, dx = \frac{1}{a} \sin ax + C$

$\int \sec^2 ax \, dx = \frac{1}{a} \tan ax + C$

$\int \csc^2 ax \, dx = -\frac{1}{a} \cot ax + C$

14. Reduction Formulas (used to simplify higher powers of trigonometric functions):

$\int \sin^n x \, dx = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2} x \, dx$

$\int \cos^n x \, dx = \frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n} \int \cos^{n-2} x \, dx$

Substitution using trigonometric identities is a technique used in calculus to simplify integrals that involve trigonometric functions. By substituting trigonometric identities, we can transform complex integrals into simpler forms that are easier to evaluate. Here’s an overview of how this technique works:

Contents hide

1 1. Identifying the Integral:

2 2. Choosing the Right Trigonometric Identity:

3 3. Applying the Substitution:

4 4. Simplifying the Integral:

5 5. Reverting to Original Variables:

6 6. Example:

7 Benefits of Using Trigonometric Substitution:

8 1. ∫sin⁡2x dx\int \sin^2 x \, dx

9 2. ∫cos⁡2x dx\int \cos^2 x \, dx

10 3. ∫sin⁡xcos⁡x dx\int \sin x \cos x \, dx

11 4. ∫sin⁡3x dx\int \sin^3 x \, dx

12 5. ∫cos⁡3x dx\int \cos^3 x \, dx

13 6. ∫sec⁡2x dx\int \sec^2 x \, dx

14 7. ∫csc⁡2x dx\int \csc^2 x \, dx

15 8. ∫sec⁡xtan⁡x dx\int \sec x \tan x \, dx

16 I=∫sec⁡xtan⁡x dx

17 ∫csc⁡xcot⁡x dx\int \csc x \cot x \, dx

18 10. ∫sin⁡x cos⁡2x dx\int \sin x \, \cos^2 x \, dx

19 11. ∫cos⁡x sin⁡x dx\int \cos x \, \sin x \, dx

20 12. ∫sin⁡2x dx\int \sin 2x \, dx

21 13. ∫cos⁡2x dx\int \cos 2x \, dx

22 14. ∫sec⁡2x tan⁡x dx\int \sec^2 x \, \tan x \, dx

23 15. ∫csc⁡2x cot⁡x dx\int \csc^2 x \, \cot x \, dx

24 16. ∫sin⁡x cos⁡2x dx\int \sin x \, \cos 2x \, dx

25 17. ∫cos⁡x sec⁡x dx\int \cos x \, \sec x \, dx

26 18. ∫sin⁡x sec⁡2x dx\int \sin x \, \sec^2 x \, dx

27 19. ∫sin⁡2xcos⁡x dx\int \sin^2 x \cos x \, dx

28 20. ∫sin⁡xcos⁡2x dx\int \frac{\sin x}{\cos^2 x} \, dx

29 21. ∫cos⁡xsin⁡3x dx\int \cos x \sin 3x \, dx

30 Also Read-

1. Identifying the Integral:

When faced with an integral involving trigonometric functions, look for opportunities to use trigonometric identities to simplify the integrand (the function being integrated). Common scenarios where substitution is useful include integrals with products of trigonometric functions or trigonometric functions raised to a power.

2. Choosing the Right Trigonometric Identity:

Select the appropriate trigonometric identity to transform the integral into a more manageable form. Common identities used in substitution include:

Pythagorean Identities:

$\sin^2 x + \cos^2 x = 1$

Double Angle Identities:

$\sin 2x = 2 \sin x \cos x$ and

$\cos 2x = \cos^2 x – \sin^2 x$

Half Angle Identities:

$\sin^2 x = \frac{1 – \cos 2x}{2}$ and

$\cos^2 x = \frac{1 + \cos 2x}{2}$

3. Applying the Substitution:

Substitute the chosen identity into the integral. This often involves expressing one trigonometric function in terms of another or converting a function to its algebraic equivalent. For example:

Example : To integrate

$\int \sin^2 x \, dx$

use the identity,

$\sin^2 x = \frac{1 – \cos 2x}{2}$

$\int \sin^2 x \, dx = \int \frac{1 – \cos 2x}{2} \, dx = \frac{1}{2} \int (1 – \cos 2x) \, dx$

\int \sin x \cos x \, dx = \frac{1}{2} \int \sin 2x \, dx

4. Simplifying the Integral:

After substitution, simplify the integral and integrate the resulting expression. The simplification usually involves straightforward algebraic manipulation or integration of elementary functions.

5. Reverting to Original Variables:

If the substitution involved a variable change (e.g., u

$u = \cos x$ replace the substituted variables with the original variables to express the final answer in terms of the original variable.

6. Example:

Consider the integral

$\int \frac{1}{\cos^2 x} \, dx$ Recognize that

$\frac{1}{\cos^2 x} = \sec^2 x$

Use the identity

$\int \sec^2 x \, dx = \tan x + C$

Steps:

Substitute

$\sec^2 x$

Integrate to get

$\tan x + C$

Benefits of Using Trigonometric Substitution:

Simplifies complex integrals.
Converts trigonometric

Substitution using trigonometric identities is a technique used in calculus to simplify integrals involving trigonometric functions. By applying specific trigonometric identities, we can transform complex integrals into simpler forms that are easier to evaluate. Here’s an overview of the process:

Here are 21 example questions involving trigonometric identities with step-by-step solutions. These examples are designed to showcase different integration techniques using trigonometric identities.

1. $\int \sin^2 x \, dx$

Solution: Use the identity,

$\sin^2 x = \frac{1 – \cos 2x}{2}$

$\int \sin^2 x \, dx = \int \frac{1 – \cos 2x}{2} \, dx$

$= \frac{1}{2} \int (1 – \cos 2x) \, dx$

$= \frac{1}{2} \left( \int 1 \, dx – \int \cos 2x \, dx \right)$

$= \frac{1}{2} \left( x – \frac{1}{2} \sin 2x \right) + C$

$= \frac{x}{2} – \frac{1}{4} \sin 2x + C$

2. $\int \cos^2 x \, dx$

Solution: Use the identity,

$\cos^2 x = \frac{1 + \cos 2x}{2}$

$\int \cos^2 x \, dx = \int \frac{1 + \cos 2x}{2} \, dx$

$= \frac{1}{2} \int (1 + \cos 2x) \, dx$

$= \frac{1}{2} \left( \int 1 \, dx + \int \cos 2x \, dx \right)$

$= \frac{1}{2} \left( x + \frac{1}{2} \sin 2x \right) + C$

$= \frac{x}{2} + \frac{1}{4} \sin 2x + C$

3. $\int \sin x \cos x \, dx$

Solution: Use the double angle identity,

$\sin 2x = 2 \sin x \cos x$ :

$\int \sin x \cos x \, dx = \frac{1}{2} \int \sin 2x \, dx$

$= -\frac{1}{2} \cos 2x + C$

4. $\int \sin^3 x \, dx$

Solution: Use,

$\sin^3 x = \sin x (1 – \cos^2 x)$

$\int \sin^3 x \, dx = \int \sin x (1 – \cos^2 x) \, dx$ Let,

$u = \cos x$ so,

$du = -\sin x \, dx$

$= -\int (1 – u^2) \, du$

$= – \left( u – \frac{u^3}{3} \right) + C$

$= – \cos x + \frac{\cos^3 x}{3} + C$

5. $\int \cos^3 x \, dx$

Solution: Use,

$\cos^3 x = \cos x (1 – \sin^2 x)$

$\int \cos^3 x \, dx = \int \cos x (1 – \sin^2 x) \, dx$ Let,

$u = \sin x$ so,

$du = \cos x \, dx$

$= \int (1 – u^2) \, du$

$= u – \frac{u^3}{3} + C$

$= \sin x – \frac{\sin^3 x}{3} + C$

6. $\int \sec^2 x \, dx$

Solution:

$\int \sec^2 x \, dx = \tan x + C$

7. $\int \csc^2 x \, dx$

Solution:

$\int \csc^2 x \, dx = -\cot x + C$

8. $\int \sec x \tan x \, dx$

Solution: Let

$u = \sec x$ so,

$d u = \sec x \tan x d x$

I=

$I=\int \sec x \tan x d x$

\int \sec x \tan x \, dx = \int du

$= u + C$

$= \sec x + C$

$\int \csc x \cot x \, dx$

Solution: Let

$u = \csc x$ , so,

$du = -\csc x \cot x \, dx$ $\int \csc x \cot x d x = - \int d u$ $= -u + C$

$= -\csc x + C$

$\int \csc x \cot x \, dx = -\int du$

10. $\int \sin x \, \cos^2 x \, dx$

Solution: Use,

$\cos^2 x = 1 – \sin^2 x$

$\int \sin x \, \cos^2 x \, dx = \int \sin x (1 – \sin^2 x) \, dx$ Let

$u = \sin x$ so,

$du = \cos x \, dx$

$= \int (u – u^3) \, du$

$= \frac{u^2}{2} – \frac{u^4}{4} + C$

$= \frac{\sin^2 x}{2} – \frac{\sin^4 x}{4} + C$

11. $\int \cos x \, \sin x \, dx$

Solution: Use the identity,

$\sin 2x = 2 \sin x \cos x$

$\int \cos x \, \sin x \, dx = \frac{1}{2} \int \sin 2x \, dx$

$= -\frac{1}{2} \cos 2x + C$

12. $\int \sin 2x \, dx$

Solution:

$\int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C$

13. $\int \cos 2x \, dx$

Solution:

$\int \cos 2x \, dx = \frac{1}{2} \sin 2x + C$

14. $\int \sec^2 x \, \tan x \, dx$

Solution: Let,

$u = \tan x$ so,

$du = \sec^2 x \, dx$

$\int \sec^2 x \, \tan x \, dx = \int u \, du$

$= \frac{u^2}{2} + C$

$= \frac{\tan^2 x}{2} + C$

15. $\int \csc^2 x \, \cot x \, dx$

Solution: Let,

$u = \cot x$ so,

$du = -\csc^2 x \, dx$

$\int \csc^2 x \, \cot x \, dx = -\int u \, du$

$= -\frac{u^2}{2} + C$

$= -\frac{\cot^2 x}{2} + C$

16. $\int \sin x \, \cos 2x \, dx$

Solution: Use the identity,

$\cos 2x = 2 \cos^2 x – 1$

$\int \sin x \, \cos 2x \, dx = \int \sin x (2 \cos^2 x – 1) \, dx$

$= 2 \int \sin x \cos^2 x \, dx – \int \sin x \, dx$ Use,

$\cos^2 x = 1 – \sin^2 x$ ;for the first integral:

$= 2 \left( \frac{\sin^2 x}{2} – \frac{\sin^4 x}{4} \right) – (-\cos x)$

$= \sin^2 x – \frac{\sin^4 x}{2} + \cos x + C$

17. $\int \cos x \, \sec x \, dx$

Solution:

$\int \cos x \, \sec x \, dx = \int \cos x \cdot \frac{1}{\cos x} \, dx = \int 1 \, dx$

$= x + C$

18. $\int \sin x \, \sec^2 x \, dx$

Solution: Use the identity,

$\sec^2 x = 1 + \tan^2 x$

$\int \sin x \, \sec^2 x \, dx = \int \sin x (1 + \tan^2 x) \, dx$

$= \int \sin x \, dx + \int \sin x \, \tan^2 x \, dx$

Let,

$u = \tan x$ so,

$du = \sec^2 x \, dx$

$= -\cos x + \int \sin x \cdot \frac{u^2}{\sec^2 x} \, du$

$= -\cos x + \int \frac{\sin x}{\cos^2 x} \, du$

$= -\cos x + \int \frac{\sin x}{1 + \tan^2 x} \, du$

$= -\cos x + \int \frac{1}{\cos x} \, du$

19. $\int \sin^2 x \cos x \, dx$

Solution: Use,

$\sin^2 x = 1 – \cos^2 x$

$\int \sin^2 x \cos x \, dx = \int (1 – \cos^2 x) \cos x \, dx$

$= \int \cos x \, dx – \int \cos^3 x \, dx$

The first integral is:

$\sin x$

$\int \cos x \, dx = \sin x$

Use

$\cos^3 x = \cos x (1 – \sin^2 x)$ : for the second integral:

$\int \cos^3 x \, dx = \int \cos x – \cos x \sin^2 x \, dx$

$= \sin x – \left( \frac{\sin^3 x}{3} \right)$

$= \sin x – \frac{\sin^3 x}{3} + C$

20. $\int \frac{\sin x}{\cos^2 x} \, dx$

Solution: Use the identity,

$\sec^2 x = \frac{1}{\cos^2 x}$

$\int \frac{\sin x}{\cos^2 x} \, dx = \int \sin x \sec^2 x \, dx$

Let

$u = \cos x$ so,

$du = -\sin x \, dx$

$= -\int \frac{1}{u^2} \, du$

$= \frac{1}{u} + C$

$= \frac{1}{\cos x} + C$

21. $\int \cos x \sin 3x \, dx$

Solution: Use the product-to-sum identity:

$\cos x \sin 3x = \frac{1}{2} [\sin (3x + x) – \sin (3x – x)]$

$= \frac{1}{2} [\sin 4x – \sin 2x]$

$\int \cos x \sin 3x \, dx = \frac{1}{2} \left( \int \sin 4x \, dx – \int \sin 2x \, dx \right)$

$= \frac{1}{2} \left( -\frac{1}{4} \cos 4x + \frac{1}{2} \cos 2x \right) + C$

$= -\frac{1}{8} \cos 4x + \frac{1}{4} \cos 2x + C$

Also Read-

https://pcmconceptclear.com/integration-by-substitution-maths12/

https://pcmconceptclear.com/integration-basic-concept-and-formula/

Integration using Trigonometric Identities Maths12: Imp Questions 2024

1. Identifying the Integral:

2. Choosing the Right Trigonometric Identity:

3. Applying the Substitution:

4. Simplifying the Integral:

5. Reverting to Original Variables:

6. Example:

Benefits of Using Trigonometric Substitution:

1. $\int \sin^2 x \, dx$

2. $\int \cos^2 x \, dx$

3. $\int \sin x \cos x \, dx$

4. $\int \sin^3 x \, dx$

5. $\int \cos^3 x \, dx$

6. $\int \sec^2 x \, dx$

7. $\int \csc^2 x \, dx$

8. $\int \sec x \tan x \, dx$

$I=\int \sec x \tan x d x$

$\int \csc x \cot x \, dx$

10. $\int \sin x \, \cos^2 x \, dx$

11. $\int \cos x \, \sin x \, dx$

12. $\int \sin 2x \, dx$

13. $\int \cos 2x \, dx$

14. $\int \sec^2 x \, \tan x \, dx$

15. $\int \csc^2 x \, \cot x \, dx$

16. $\int \sin x \, \cos 2x \, dx$

17. $\int \cos x \, \sec x \, dx$

18. $\int \sin x \, \sec^2 x \, dx$

19. $\int \sin^2 x \cos x \, dx$

20. $\int \frac{\sin x}{\cos^2 x} \, dx$

21. $\int \cos x \sin 3x \, dx$

Also Read-

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