Let k be an integer greater or equal to zero, show that
$$
\int_0^{\pi } \cos ^{2 k+1}(x) \, dx=0 \
\int_0^{\pi } \cos ^{2 k}(x) \, dx=\frac{\pi (2
k)!}{2^{2 k} (k!)^2}
$$

Question

Let k be an integer greater or equal to zero, show that
$$
\int_0^{\pi } \cos ^{2 k+1}(x) \, dx=0 \
\int_0^{\pi } \cos ^{2 k}(x) \, dx=\frac{\pi  (2
   k)!}{2^{2 k} (k!)^2}
$$

anonymous · Answer

Hint, read http://en.wikipedia.org/wiki/Integration_by_reduction_formulae

anonymous · Answer

$$Let f(x)=\left( \cos x \right)^{2k+1}$$
$$f \left( \pi-x \right)=\left\{ \cos \left( \pi-x \right) \right\}^{2k-1}=\left( -\cos x \right)^{2k+1}=-\left( \cos x \right) ^{2k+1}=-f \left( x \right)$$
$$\int\limits_{0}^{\pi}f(x)=0$$

ganeshie8 · Answer

$$\begin{align}\int_0^{\pi } \cos ^{2 k}(x) \, dx&= \frac{2k-1}{2k} \int_0^{\pi} \cos^{2k-2}(x) \,dx\~\
&=\frac{2k-1}{2k} \cdot \frac{2k-3}{2k-2}\cdots \int_0^{\pi} 1\, dx

\end{align}$$

anonymous · Answer

http://openstudy.com/users/eliassaab#/updates/54a358eae4b0b8a54fb07d0b#/updates/54a358eae4b0b8a54fb07d0b

anonymous · Answer

This result was used to prove the link above. 
@SithsAndGiggles