Question

Evaluate : $\int sinxcos\frac{x}{2}cos\frac{3x}{2}dx$

• A. $cosx-\frac{1}{2}cos2x+\frac{1}{3}cos3x+c$
• B. $\frac{1}{4}cosx-\frac{1}{8}cos2x-\frac{1}{12}cos3x+c$
• C. $cosx+\frac{1}{2}cos2x+\frac{1}{3}cos3x+c$
• D. $cosx+\frac{1}{2}cos2x-\frac{1}{3}cos3x+c$

Answer 1

The correct option is B $\frac{1}{4}cosx-\frac{1}{8}cos2x-\frac{1}{12}cos3x+c$

$\int \sin x\cos \frac{x}{2}\cos \frac{3x}{2}dx$

$=\frac{1}{2}\int \sin x\left(2\cos \frac{x}{2}\cos \frac{3x}{2}\right)dx$ [by Trigonometric transformation formulas]

$=\frac{1}{2}\int \sin x(\cos x+\cos 2x)dx$

$=\frac{1}{2}\int \sin x\cos xdx+\frac{1}{2}\int \sin x\cos 2xdx$

$=\frac{1}{4}\int \sin 2xdx+\frac{1}{4}\int \sin 3xdx-\frac{1}{4}\int \sin xdx$ [by Trigonometric transformation formulas]

$=-\frac{1}{8}\cos 2x-\frac{1}{12}\cos 3x+\frac{1}{4}\cos x$