Evaluate the integral int {left( {sin x cos x | vedclass

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Question

(D) Let $I = \int {\sin x \cos x \cos 2x \cos 4x \cos 8x} \, dx$. Multiply and divide by $2$: $I = \frac{1}{2} \int {\sin 2x \cos 2x \cos 4x \cos 8x}

Vedclass · Accepted Answer

$\frac{{ - 1}}{{256}}\cos 16x + C$

Vedclass · Answer

(D) Let $I = \int {\sin x \cos x \cos 2x \cos 4x \cos 8x} \, dx$. Multiply and divide by $2$: $I = \frac{1}{2} \int {\sin 2x \cos 2x \cos 4x \cos 8x} \, dx$. Multiply and divide by $2$ again: $I = \frac{1}{4} \int {\sin 4x \cos 4x \cos 8x} \, dx$. Multiply and divide by $2$ again: $I = \frac{1}{8} \int {\sin 8x \cos 8x} \, dx$. Multiply and divide by $2$ again: $I = \frac{1}{16} \int {\sin 16x} \, dx$. Integrating $\sin 16x$,we get: $I = \frac{1}{16} \left( -\frac{\cos 16x}{16} \right) + C = -\frac{1}{256} \cos 16x + C$.

Evaluate the integral $\int {\left( {\sin x \cos x \cos 2x \cos 4x \cos 8x} \right) dx}$.

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