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?
 A: Hint for question
$\int_0^{\pi}\cos^{2n} \theta d\theta=2\int_0^{\pi/2}\cos^{2n} \theta d\theta$
which is Wallis' integral: integral
Now for n=3 calculate the answer
Since $W_6=\frac{5\pi}{32} \Rightarrow \int_0^{\pi}\cos^{6} \theta d\theta=\frac{5\pi}{16}$
A: Answer (using partial integration):
$$\int\cos^6(x)\space\space\text{d}x=$$
$$\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{6}\int\cos^4(x)\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{8}\int\cos^2(x)\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{8}\int\left(\frac{1}{2}\cos(2x)+\frac{1}{2}\right)\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{32}\int\cos(2x)\space\space\text{d}x+\frac{5}{16}\int 1\space\space\text{d}x=$$

Substitute $u=2x$ and $\text{d}u=2\space\space\text{d}x$:

$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{32}\int\cos(u)\space\space\text{d}u+\frac{5}{16}\int 1\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5\sin\left(u\right)}{32}+\frac{5x}{16}+\text{C}=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5\sin\left(2x\right)}{32}+\frac{5x}{16}+\text{C}=$$
$$\frac{1}{192}\left(60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right)+\text{C}$$

With boundaries:
$$\left[\frac{1}{192}\left(60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right)\right]_{0}^{\pi}=$$
$$\frac{1}{192}\left[60x+45\sin(2x)+9\sin(4x)+\sin(6x)\right]_{0}^{\pi}=$$
$$\frac{1}{192}\left(60\pi+45\sin(2\pi)+9\sin(4\pi)+\sin(6\pi)\right)=$$
$$\frac{1}{192}\left(60\pi+0+0+0\right)=\frac{60\pi}{192}=\frac{5\pi}{16}$$
A: $$\int^\pi_0\cos^6\theta\ d\theta=\int^\pi_0(\frac{1+\cos2\theta}{2})^3 d\theta$$
$$\frac{1}{8}\int^\pi_0(1+\cos2 \theta)^3d\theta=\frac{1}{8}\int^\pi_0(1+3\cos2\theta+3\cos^2 2\theta+\cos^32\theta)d\theta $$
$$\frac{1}{8}\int^\pi_0(1+3\cos2\theta+\frac{3}{2}(1+cos 4\theta)+\cos2\theta(1-\sin^22\theta)))d\theta$$ 
see, all terms can be integrated directly
A: 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,
$$\int_0^\pi \cos^6\theta d\theta = \frac{1}{2^6} \binom{6}{3}\pi = \frac{5}{16}\pi.$$
