How to calculate $\frac{I_{n+2}}{I_n}$ of $I_n = \int_{\frac {-\pi}{2}}^\frac{\pi}{2} cos^n \theta d\theta$

Question

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...

Answer 1

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 $\cos\pm\frac{\pi}{2}=0$.

Answer 2

\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}

Answer 3

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 $$\int_0^{\pi/2}\sin^at\cos^bt\ dt=\int_0^{\pi/2}\sin^bt\cos^at\ dt$$

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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: $$\int_0^{\pi/2}\tan^at\ dt=\frac{\pi}2\csc\frac{\pi(a+1)}2=\int_0^{\pi/2}\cot^at\ dt$$