Question

$\int 4\sin xcos\frac{x}{2}\cos \frac{3x}{2}dx$ equals :

• A. $\cos x-\frac{1}{2}\cos 2x+\frac{1}{3}\cos 3x+c$
• B. $\cos x-\frac{1}{2}\cos 2x-\frac{1}{3}\cos 3x+c$
• C. $\cos x+\frac{1}{2}\cos 2x+\frac{1}{3}\cos 3x+c$
• D. $\cos x+\frac{1}{2}\cos 2x-\frac{1}{3}\cos 3x+c$

Answer 1

The correct option is B $\cos x-\frac{1}{2}\cos 2x-\frac{1}{3}\cos 3x+c$

Consider, $I=\int 4\sin x\cos \frac{x}{2}\cos \frac{3x}{2}dx$

$\Rightarrow \int 2\cos A\cos B=\cos (A+B)+\cos (A-B)$

$\Rightarrow \int 2\sin x[\cos 2x+\cos x]dx$

$\Rightarrow \int 2\sin x\cos xdx+2\int 2\sin x\cos xdx$

$\Rightarrow 2\sin A\cos 2B=\sin (A+B)+\sin (A-B)$

$\Rightarrow \int (\sin 3x-\sin x)dx+\int \sin 2xdx$

$\Rightarrow \frac{-\cos 3x}{3}+\cos x-\frac{\cos 2x}{2}+c$