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CBSE Class 12 · 2007 · Indefinite IntegralsLong Answer

Evaluate :

\int sin 4 x sin 8 x d x

Answer: $\frac{1}{8} sin 4 x - \frac{1}{24} sin 12 x + C$

Solution

Convert the product to a difference using

2 sin A sin B = cos (A - B) - cos (A + B)

:

sin 4 x sin 8 x = \frac{1}{2} (cos 4 x - cos 12 x) .

Integrating,

\frac{1}{2} \int (cos 4 x - cos 12 x) d x = \frac{1}{8} sin 4 x - \frac{1}{24} sin 12 x + C

CBSE Class 12 Board · 2007 · Subjective

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