IIT JEE Main Maths -Unit 8- Integration using trigonometric identities- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 8- Integration using trigonometric identities – Study Notes – New syllabus
IIT JEE Main Maths -Unit 8- Integration using trigonometric identities – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Integration of Trigonometric Functions (Using Identities and Substitutions)
Integration of Trigonometric Functions (Using Identities and Substitutions)
Many integrals involving trigonometric functions can be simplified using standard identities and substitutions. These techniques help convert complex trigonometric expressions into standard integrable forms.
Basic Trigonometric Identities Used
- \( \sin^2x + \cos^2x = 1 \)
- \( 1 + \tan^2x = \sec^2x \)
- \( 1 + \cot^2x = \csc^2x \)
- \( 2\sin A \cos B = \sin(A + B) + \sin(A – B) \)
- \( 2\cos A \cos B = \cos(A + B) + \cos(A – B) \)
- \( 2\sin A \sin B = \cos(A – B) – \cos(A + B) \)
Standard Integrals of Basic Trigonometric Functions
| Function \( f(x) \) | Integral \( \displaystyle \int f(x)\,dx \) |
|---|---|
| \( \sin x \) | \( -\cos x + C \) |
| \( \cos x \) | \( \sin x + C \) |
| \( \sec^2 x \) | \( \tan x + C \) |
| \( \csc^2 x \) | \( -\cot x + C \) |
| \( \sec x \tan x \) | \( \sec x + C \) |
| \( \csc x \cot x \) | \( -\csc x + C \) |
Reduction of Powers (using identities)
These formulas are useful when dealing with powers of sine and cosine:
- \( \sin^2x = \dfrac{1 – \cos 2x}{2} \)
- \( \cos^2x = \dfrac{1 + \cos 2x}{2} \)
- \( \sin x \cos x = \dfrac{1}{2}\sin 2x \)
Example
Evaluate \( \displaystyle \int \sin^2x\,dx \).
▶️ Answer / Explanation
Step 1: Use identity \( \sin^2x = \dfrac{1 – \cos 2x}{2} \).
Step 2: Substitute and integrate:
\( \displaystyle \int \sin^2x\,dx = \dfrac{1}{2}\int (1 – \cos 2x)\,dx = \dfrac{x}{2} – \dfrac{\sin 2x}{4} + C. \)
Answer: \( \dfrac{x}{2} – \dfrac{\sin 2x}{4} + C \).
Example
Evaluate \( \displaystyle \int \sin 3x \cos 2x\,dx \).
▶️ Answer / Explanation
Step 1: Use the product-to-sum formula: \( 2\sin A \cos B = \sin(A + B) + \sin(A – B) \).
So, \( \sin 3x \cos 2x = \dfrac{1}{2}[\sin(5x) + \sin x] \).
Step 2: Substitute and integrate:
\( \displaystyle \int \sin 3x \cos 2x\,dx = \dfrac{1}{2}\int (\sin 5x + \sin x)\,dx \)
\( = \dfrac{1}{2}\left(-\dfrac{\cos 5x}{5} – \cos x\right) + C. \)
Answer: \( -\dfrac{1}{10}\cos 5x – \dfrac{1}{2}\cos x + C. \)
Example
Evaluate \( \displaystyle \int \dfrac{dx}{1 + \sin x} \).
▶️ Answer / Explanation
Step 1: Multiply numerator and denominator by \( 1 – \sin x \):
\( \displaystyle \int \dfrac{dx}{1 + \sin x} = \int \dfrac{1 – \sin x}{\cos^2x}\,dx = \int \sec^2x\,dx – \int \tan x \sec x\,dx. \)
Step 2: Integrate separately: \( \int \sec^2x\,dx = \tan x, \quad \int \tan x \sec x\,dx = \sec x. \)
Step 3: Combine results: \( \tan x – \sec x + C. \)
Answer: \( \tan x – \sec x + C. \)
Common Substitutions
| Expression | Substitute |
|---|---|
| \( \sin x, \cos x \) | Use half-angle formulas |
| \( \tan x \text{ or } \sec x \) | Let \( t = \tan\dfrac{x}{2} \) (Weierstrass Substitution) |
| \( \sin x, \cos x \) in rational form | Let \( \tan\dfrac{x}{2} = t \), use \( \sin x = \dfrac{2t}{1 + t^2}, \, \cos x = \dfrac{1 – t^2}{1 + t^2}, \, dx = \dfrac{2\,dt}{1 + t^2} \) |