$\int_{0}^{π} \cos 6 x \cdot \cos^{8} xdx =$

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$\frac{π}{16}$

$\frac{π}{64}$

$\frac{π}{32}$

$\frac{π}{8}$

answer is C.

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Detailed Solution

$\cos^{8} x = \frac{1}{2^{8}} {(e^{i x} + e^{- i x})}^{8} = \frac{1}{2^{8}} ((e^{i 8 x} + e^{- i 8 x}) + 8 (e^{6 ix} + e^{- i 6 x}) + 28 (e^{i 4 x} + e^{- i 4 x}) + 56 (e^{i 2 x} + e^{- i 2 x}) + 70) = \frac{1}{2^{7}} (\cos 8 x + 8 \cos 6 x + 28 \cos 4 x + 56 \cos 2 x + 35)$
Using binomial Expansion $(x + y)^{n} =^{n} C_{0} x^{n} +^{n} C_{1} x^{n - 1} y^{1} + \dots . .]$
$\begin{array}{l} = \frac{1}{2^{7}} (\cos 8 x + 8 \cos 6 x + \dots \dots) \\ \begin{aligned} \int_{0}^{π} \cos 6 x \cdot \cos^{8} xdx = \int_{0}^{π} \cos 6 x \cdot \frac{8}{2} \cos 6 xdx \\ = \frac{1}{16} \int_{0}^{π} \cos^{2} 6 xdx = \int_{0}^{π} \frac{1 + \cos 12 x}{32} dx = \frac{π}{32} \end{aligned} \end{array}$