Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 20  Maths Textbook Solution.

$-\frac{\cos 4 x}{16}-\frac{\cos 2 x}{8}+\frac{\cos 6 x}{24}+c$

Given:

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

Hint:

Use the formula $2 \sin A \sin B=-\cos (A+B)+\cos (A-B)\rfloor$

Solution:

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

$=\int \frac{1}{2}[\cos (x-2 x)-\cos (x+2 x)] \sin 3 x d x$

$=\int \frac{1}{2}(\cos (-\mathrm{x})-\cos 3 \mathrm{x}) \sin 3 \mathrm{x} \mathrm{dx}$

$=\frac{1}{2} \int \cos x \sin 3 x-\cos 3 x \sin 3 x d x$

$=\frac{1}{2} \int \frac{1}{2}(\sin 4 x+\sin 2 x) d x-\frac{1}{2} \int \frac{1}{2}(\sin 6 x) d x$

$=\frac{1}{4}\left[\int(\sin 4 x+\sin 2 x-\sin 6 \mathrm{x}) \mathrm{dx}\right]$

$=\frac{1}{4}\left[-\frac{\cos 4 x}{4}-\frac{\cos 2 x}{2}+\frac{\cos 6 x}{6}\right]+c$

$-\frac{\cos 4 x}{16}-\frac{\cos 2 x}{8}+\frac{\cos 6 \mathrm{x}}{24}+\mathrm{c}$

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