CBSE Class 12 · 2012 · Indefinite IntegralsLong Answer

Evaluate :

\int sin x sin 2 x sin 3 x d x

Answer: $- \frac{1}{8} cos 2 x + \frac{1}{24} cos 6 x - \frac{1}{16} cos 4 x + C$

Solution

First combine two factors:

2 sin x sin 3 x = cos 2 x - cos 4 x

, so

sin x sin 2 x sin 3 x = \frac{1}{2} sin 2 x (cos 2 x - cos 4 x) .

Now

\frac{1}{2} sin 2 x cos 2 x = \frac{1}{4} sin 4 x

and

2 sin 2 x cos 4 x = sin 6 x - sin 2 x

, so the integrand becomes

\frac{1}{4} sin 4 x - \frac{1}{4} (sin 6 x - sin 2 x) .

Integrating term by term,

- \frac{1}{8} cos 2 x + \frac{1}{24} cos 6 x - \frac{1}{16} cos 4 x + C

CBSE Class 12 Board · 2012 · Subjective