Question

The value of $\int \left(\sin x\cos x\cos 2x\cos 4x\cos 8x\cos 16x\right)dx$ is equal to

• A. $\frac{\sin 16x}{1024}+C$
• B. $-\frac{\cos 32x}{1024}+C$
• C. $\frac{\cos 32x}{1096}+C$
• D. $-\frac{\cos 32x}{1096}+C$

Answer 1

Answer: (B)

$-\frac{\cos 32x}{1024}+C$

$I=\int \left(\sin x\cos x\cos 2x\cos 4x\cos 8x\cos 16x\right)dx$

$I=\frac{1}{2}\int \left(2\sin x\cos x\right)\left(\cos 2x\cos 4x\cos 8x\cos 16x\right)dx$

$=\frac{1}{2}\int \left(\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x\right)dx$

$=\frac{1}{4}\int \left(2\sin 2x\cos 2x\right)\left(\cos 4x\cos 8x\cos 16x\right)dx$

$=\frac{1}{4}\int \left(\cos 4x\sin 4x\right)\left(\cos 8x\cos 16x\right)dx$

$=\frac{1}{8}\int \left(\sin 8x\right)\left(\cos 8x\right)\left(\cos 16x\right)dx$

$=\frac{1}{16}\int \sin 16x\cos 16xdx$

$I=\frac{1}{32}$ $\int \sin \left(32x\right)dx$

put $\cos \left(32x\right)=t$

$-\left(\sin \right(32x\left)\right)\times 32dx=dt$

$I=\frac{1}{32}\times (-\frac{1}{32})\int dt$

$=-\frac{1}{(32{)}^2}t+c$

$I=\frac{-\cos \left(32x\right)}{1024}+c$

$\therefore \int \sin x\cos x\cos 2x\cos 4x\cos 8x\cos 16xdx=\frac{-\cos 32x}{1024}+c$.