How to calculate $\frac{I_{n+2}}{I_n}$ of $I_n = \int_{\frac {-\pi}{2}}^\frac{\pi}{2} cos^n \theta d\theta$ How do I calculate the $\frac{I_{n+2}}{I_n}$ of $I_n = \int_{\frac {-\pi}{2}}^\frac{\pi}{2} cos^n \theta d\theta$ ?
[my attempt]:
I could calculate that $nI_n = 2cos^{n-1}\theta sin\theta+2(n-1)\int_0^\frac{\pi}{2}I_{n-2}d\theta$ but how do I caluclate the $I_{n+2}$? 
I am stacking there...
 A: Correction: you meant $nI_n=[\cos^{n-1}\theta\sin\theta]_{-\pi/2}^{\pi/2}+(n-1)(I_{n-2}-I_n)$. The first term vanishes if $n\ge 2$ because $\cos\pm\frac{\pi}{2}=0$.
A: \begin{align}
I_n
&= \int_{-\pi/2}^{\pi/2}\cos^n\theta\ d\theta \\
&= 2\int_{0}^{\pi/2}\cos^n\theta\ d\theta \\
&= {\bf B}\left(\dfrac{1}{2},\dfrac{n+1}{2}\right) \\
&= \dfrac{\Gamma\left(\dfrac{1}{2}\right)\Gamma\left(\dfrac{n+1}{2}\right)}{\Gamma\left(\dfrac{n}{2}+1\right)} \\
\dfrac{I_{n+2}}{I_{n}}
&= \dfrac{\Gamma\left(\dfrac{1}{2}\right)\Gamma\left(\dfrac{n+3}{2}\right)}{\Gamma\left(\dfrac{n+2}{2}+1\right)} 
\dfrac{\Gamma\left(\dfrac{n}{2}+1\right)}{\Gamma\left(\dfrac{1}{2}\right)\Gamma\left(\dfrac{n+1}{2}\right)} \\
&= \color{blue}{\dfrac{n+1}{n+2}}
\end{align}
A: Expanding on Nosrati's answer:
Consider the integral 
$$I(a,b)=\int_0^{\pi/2}\sin^at\cos^bt\ dt$$
The substitution $x=\sin^2t$ gives
$$I(a,b)=\frac12\int_0^1x^{\frac{a-1}2}(1-x)^{\frac{b-1}2}dx$$
$$I(a,b)=\frac12\int_0^1x^{\frac{a+1}2-1}(1-x)^{\frac{b+1}2-1}dx$$
Recall the definition of the Beta Function:
$$B(u,v)=\int_0^1t^{u-1}(1-t)^{v-1}dt$$
This gives our integral:
$$I(a,b)=\frac12B\bigg(\frac{a+1}2,\frac{b+1}2\bigg)$$
We may recall that 
$$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
Where $\Gamma(\cdot)$ is the Gamma Function. This gives
$$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}=\frac{\Gamma(\frac{b+1}2)\Gamma(\frac{a+1}2)}{2\Gamma(\frac{a+b}2+1)}=I(b,a)$$
Thus we have proven the interesting identity
$$\int_0^{\pi/2}\sin^at\cos^bt\ dt=\int_0^{\pi/2}\sin^bt\cos^at\ dt$$

Extra fun on the side:
Noting that $\tan t=\sin t (\cos t)^{-1}$ gives
$$Q=\int_0^{\pi/2}\tan^at\ dt=I(a,-a)=\frac{\Gamma(\frac{1+a}2)\Gamma(\frac{1-a}2)}{2\Gamma(1)}\\Q=\frac12\Gamma\bigg(\frac{1+a}2\bigg)\Gamma\bigg(\frac{1-a}2\bigg)$$
We may recall the Gamma reflection formula: for non-integer $x$,
$$\Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)$$
Plugging in $x=\frac{a+1}2$ gives 
$$Q=\frac{\pi}2\csc\frac{\pi(a+1)}2$$
Hence our conclusion:
$$\int_0^{\pi/2}\tan^at\ dt=\frac{\pi}2\csc\frac{\pi(a+1)}2=\int_0^{\pi/2}\cot^at\ dt$$
