Question

$\int \sin x.\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16xdx$ is equal to

• A. $\frac{\sin 16x}{1024}+c$
• B. $-\frac{\cos 32x}{1024}+c$
• C. $\frac{\cos 32x}{1096}+c$
• D. none

Answer 1

Answer: (B)

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

$\int \sin x.\cos x.\cos 2x.\cos 4x.cos8x.\cos 16xdx$

$\Rightarrow$ $\frac{1}{2}\int (2\sin x.\cos x)cos2x.\cos 4x.cos8x.\cos 16xdx$

$\Rightarrow$ $\frac{1}{2^2}\int (2\sin 2x.\cos 2x)\cos 4x.cos8x.\cos 16xdx$

$\Rightarrow$ $\frac{1}{2^3}\int (2\sin 4x.\cos 4x).\cos 8x.\cos 16xdx$

$\Rightarrow$ $\frac{1}{2^4}\int (2\sin 8x.\cos 8x).\cos 16xdx$

$\Rightarrow$ $\frac{1}{2^5}\int \left(2\sin 16x\cos 16x\right)dx$

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

$\Rightarrow$ $\frac{1}{32}\times \frac{(-cos32x)}{32}+c$

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