Definite Integration
Integration of trigonometric functions
Grade Class 12
Question:
15. $\int [\sin \alpha \sin(x - \alpha) + \sin^2(\frac{x}{2} - \alpha)] dx$ equals
(A) $\frac{1}{2}(x + \sin x) + C$
(B) $\frac{1}{2}(x^2 - \sin x) + C$
(C) $\frac{1}{2}(x - \sin x) + C$
(D) $\frac{1}{2}(x - \cos x) + C$
Step-by-Step Solution
Key Concept: Use trigonometric identities: 2sin A sin B = cos(A-B) - cos(A+B) and 2sin^2(theta) = 1 - cos(2theta).
The integral is $\int [\sin \alpha \sin(x - \alpha) + \sin^2(\frac{x}{2} - \alpha)] dx$. Using $2\sin A \sin B = \cos(A-B) - \cos(A+B)$, we have $\sin \alpha \sin(x - \alpha) = \frac{1}{2}[\cos(x - 2\alpha) - \cos(x)]$. Using $2\sin^2 \theta = 1 - \cos 2\theta$, we have $\sin^2(\frac{x}{2} - \alpha) = \frac{1}{2}[1 - \cos(x - 2\alpha)]$. Adding these, the integral becomes $\int [\frac{1}{2}\cos(x - 2\alpha) - \frac{1}{2}\cos x + \frac{1}{2} - \frac{1}{2}\cos(x - 2\alpha)] dx = \int [\frac{1}{2} - \frac{1}{2}\cos x] dx = \frac{1}{2}x - \frac{1}{2}\sin x + C = \frac{1}{2}(x - \sin x) + C$.
Correct Answer: 3