Question

The value of $\int 8\cos \left(\frac{5x}{2}\right)\cdot \sin 2x\cdot \cos \left(\frac{3x}{2}\right)dx$ is
(where $C$ is constant of integration)

• A. $-2\cos x+\frac{\cos 2x}{2}-\frac{2\cos 3x}{3}-\frac{\cos 6x}{6}+C$
• B. $2\cos x+\frac{\cos 2x}{2}+\frac{2\cos 3x}{3}-\frac{\cos 6x}{6}+C$
• C. $-2\cos x+\cos 2x-\frac{2\cos 3x}{3}-\frac{\cos 6x}{3}+C$
• D. $2\cos x+\cos 2x-\frac{2\cos 3x}{3}-\frac{\cos 6x}{6}+C$

Answer 1

The correct option is C $-2\cos x+\cos 2x-\frac{2\cos 3x}{3}-\frac{\cos 6x}{3}+C$

Let $I=\int 8\cos \left(\frac{5x}{2}\right)\cdot \sin 2x\cdot \cos \left(\frac{3x}{2}\right)dx$
$\Rightarrow I=\int \left(4\sin 2x\right[\cos 4x+\cos x\left]\right)dx$
$(\because 2\cos a\cos b=\cos (a+b)+\cos (a-b\left)\right)$
$\Rightarrow I=2\int (2\sin 2x\cos 4x+2\sin 2x\cos x])dx$
$\Rightarrow I=2\int \left(\right(\sin 6x+\sin (-2x))+(\sin 3x+\sin x\left)\right)dx$
$\Rightarrow I=2\int \left(\right(\sin 6x-\sin 2x)+(\sin 3x+\sin x\left)\right)dx$
$\Rightarrow I=2(\frac{-\cos 6x}{6}+\frac{\cos 2x}{2}-\frac{\cos 3x}{3}-\frac{\cos x}{1})+C$
$\therefore I=-2\cos x+\cos 2x-\frac{2\cos 3x}{3}-\frac{\cos 6x}{3}+C$