Question

$\int \sin x.\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16xdx$ equals

• 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

The correct option is B $-\frac{\cos 32x}{1024}+c$

Given,
$\int \sin x.\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16xdx$

$=\int \frac{1}{2}\sin 2x.\cos 2x.\cos 4x.\cos 8x.\cos 16xdx$

$=\int \frac{1}{4}\sin 4x.\cos 4x.\cos 8x.\cos 16xdx$

$=\int \frac{1}{8}\sin 8x.\cos 8x.\cos 16xdx$

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

$=\int \frac{1}{32}\sin 32xdx$

$=-\frac{\cos 32x}{32\times 32}$+c

$=-\frac{\cos 32x}{1024}$+c