#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 21 Maths Textbook Solution.

$I=\frac{1}{4}\left(x+\frac{\sin 6 x}{6}+\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c$

Given:

$\int \cos x \cos 2 x \cos 3 x d x$

Hint:

To solve this equation  we will use trigonometry method.

Solution:

$\int(\cos x \cos 2 x) \cos 3 x d x$

$I=\int \frac{\cos (x+2 x)+\cos (x-2 x)}{2} \cos 3 x d x \quad\left[\because \cos C \cdot \cos D=\frac{1}{2} \cos (C+D)+\cos (C-D)\right)$

$I=\frac{1}{2} \int(\cos 3 x+\cos x) \cos 3 x d x$

$I=\frac{1}{2}\left(\int \cos ^{2} 3 x d x+\int \cos x \cos 3 x\right) d x$

$I=\frac{1}{2} \int \frac{1+\cos 6 x}{2} d x+\frac{1}{2} \int \cos (x+3 x)+\cos (x-2 x) d x$

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

$I=\frac{1}{2} \int \frac{1}{2} d x+\int \frac{\cos 6 x}{2} d x+\frac{1}{2} \int \cos 4 x d x+\frac{1}{2} \int \cos 2 x d x$

$I=\frac{1}{2}\left(\frac{1}{2} x+\frac{1}{2} \frac{\sin 6 x}{6}+\frac{1}{2} \frac{\sin 4 x}{4}+\frac{1}{2} \frac{\sin 2 x}{2}\right)+c$

$I=\frac{1}{2}\left(\frac{x}{2}+\frac{\sin 6 x}{12}+\frac{\sin 4 x}{8}+\frac{\sin 2 x}{4}\right)+c$