First slide

Introduction to integration

Question

$\int \sin x \cdot \cos x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 x d x is equal to$

Moderate

A

$- \frac{1}{96} \cos 16 x + C$
B

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

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

$None of these$

Solution

$\begin{array}{l} I = \int \sin x \cdot \cos x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 xdx \\ = \frac{1}{2} \int \sin 2 x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 xdx \\ = \frac{1}{4} \int \sin 4 x \cdot \cos 4 x \cdot \cos 8 xdx \\ = \frac{1}{8} \int \sin 8 x \cdot \cos 8 xdx \\ = \frac{1}{16} \int \sin 16 xdx = \frac{- 1}{256} \cos 16 x + C \end{array}$

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