<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] $$\int^\pi_0\cos^6\theta\ d\theta=\int^\pi_0(\frac{1+\cos2\theta}{2})^3 d\theta$$
$$\frac{1}{8}\int^\pi_0(1+\cos2 \theta)^3d\theta=\frac{1}{8}\int^\pi_0(1+3\cos2\theta+3\cos^2 2\theta+\cos^32\theta)d\theta $$
$$\frac{1}{8}\int^\pi_0(1+3\cos2\theta+\frac{3}{2}(1+cos 4\theta)+\cos2\theta(1-\sin^22\theta)))d\theta$$ 
see, all terms can be integrated directly
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