<s>[INST] Evaluate the Integral: $\int^\pi_0\cos^6\theta\ d\theta$ $\int^\pi_0\cos^6\theta\ d\theta$
So I split the trig value into:
$\int^\pi_o\cos^5\theta\ cos\theta\ d\theta$
Then I utilized the Pythagorean theorem for $cos^5\theta$
$\int^\pi_o(1-sin^5\theta)\ cos\theta$
I utilized u-substitution:
$u=sin\ \theta$
$du=cos\ \theta$ 
Thus:
$\int^{x=\pi}_{x=0}\ (1-u^5)\ d\theta$
I intergated
$(\frac{1}{6}u^6)+(\frac{1}{6}u^6)$
$-(\frac{\pi^6}{6})+(0)$
$-(\frac{\pi^6}{6})$
Is my answer right?
 [/INST] An easy way is to use Euler formula:
$$\cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$$
Then use the binomial expansion:
$$\cos^6\theta = \frac{1}{2^6}\sum_{k = 0}^6 \binom{6}{k}e^{i(6 - k)\theta}e^{-ik\theta} = \frac{1}{2^6}\sum_{k = 0}^6 \binom{6}{k}e^{i(6 - 2k)\theta}.$$
If $k = 3$, $\int_0^\pi e^{i(6 - 2k)\theta} d\theta = \pi.$ Otherwise
$$\int_0^\pi e^{i(6 - 2k)\theta} d\theta = \frac{1}{6 - 2k}(e^{i(6 - 2k)\pi} - 1) = 0$$
in view of $6 - 2k$ is always an even number. Therefore,
$$\int_0^\pi \cos^6\theta d\theta = \frac{1}{2^6} \binom{6}{3}\pi = \frac{5}{16}\pi.$$
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