Definite Integration
Trigonometric Identities
Grade Class 12
Question:
$\int \left[ \sin \alpha \sin(x - \alpha) + \sin^2 \left( \frac{x}{2} - \alpha \right) \right] dx$ equals
$\frac{1}{2}(x + \sin x) + C$
$\frac{1}{2}(x^2 - \sin x) + C$
$\frac{1}{2}(x - \sin x) + C$
$\frac{1}{2}(x - \cos x) + C$
Step-by-Step Solution
Key Concept: Use trigonometric identities: sin(A)sin(B) = 1/2[cos(A-B) - cos(A+B)] and sin^2(theta) = (1 - cos(2theta))/2.
The integral is $I = \int \sin \alpha \sin(x - \alpha) dx + \int \sin^2 \left( \frac{x}{2} - \alpha \right) dx$. Using $\sin \alpha \sin(x - \alpha) = \frac{1}{2} [\cos(x - 2\alpha) - \cos x]$ and $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$, we get $I = \frac{1}{2} \int (\cos(x - 2\alpha) - \cos x) dx + \frac{1}{2} \int (1 - \cos(x - 2\alpha)) dx = \frac{1}{2} \int (1 - \cos x) dx = \frac{1}{2}(x - \sin x) + C$.
Correct Answer: C