\( \int\sin 3x\,\sin 2x\,dx= \) (where c is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{2}\sin x-\displaystyle\frac{\sin 5x}{10}+c \)
The product becomes a difference by the identity \( 2\sin A\sin B=\cos(A-B)-\cos(A+B) \):
\( \displaystyle \sin 3x\sin 2x=\frac{1}{2}(\cos x-\cos 5x) \)
Integrating term by term:
\( \displaystyle \int\sin 3x\sin 2x\,dx \)
\( \displaystyle =\frac{1}{2}\left[\sin x-\frac{\sin 5x}{5}\right]+c \)
\( \displaystyle =\frac{1}{2}\sin x-\frac{\sin 5x}{10}+c \)