# integration of $\int_0^{\frac{\pi}{2}} \cos^{n}(t)dt$

$$\int_0^{\frac{\pi}{2}}\cos^{n}(t)dt =?$$ To solve this problem, I was thinking that I would let $$\cos(t)= \frac{e^{it} + e^{-it}}{2}$$, then the integral will have the form: $$\int_0^{\frac{\pi}{2}} \left (\frac{e^{it} + e^{-it}}{2} \right)^{n}dt=\frac{1}{2^{n}}\,\sum_{k=0}^{n}\binom{n}{k}\int_0^{\frac{\pi}{2}}e^{i(n-2k)t}\,dt$$

From this point, I was stuck. So, would anyone please help me to walk through this problem.

• consider integration by parts and induction May 7, 2020 at 14:29
• See Wallis' integral. The article gives a method to evaluate this. May 7, 2020 at 14:32
• This is well known Wallis' integrals May 7, 2020 at 14:32

Use integration by parts on $$I(n)=\int_0^{\frac{\pi}{2}}\cos^{n}(t)dt=\int_0^{\frac{\pi}{2}}\cos(t)\cos^{n-1}(t)dt$$ then use the pythagorean identity on $$\sin^2(t)$$. You should end up with $$I(n)=\frac{n-1}{n}\int_0^{\frac{\pi}{2}}\cos^{n-2}(t)dt=\frac{n-1}{n}I(n-2)$$

• Thank you, In this way I will solve the question. May 7, 2020 at 15:09

$$\cos^n(t)=\sin^n\left(\frac{\pi}{2}-t\right)$$

Let:

$$\frac{\pi}{2}-t=x$$,

$$t=0 \implies x=\frac{\pi}{2}$$

$$t=\frac{\pi}{2} \implies x=0$$

$$\mathrm{d}t=-\mathrm{d}x$$

$$I=\int^{\frac{\pi}{2}}_0 \cos^n(t)~\mathrm{d}t=-\int^{\frac{\pi}{2}}_0 \sin^n(x)\mathrm{d}x=-\int^{\frac{\pi}{2}}_0 \sin (x)(1-\cos^2x)^{\frac{n-1}{2}}\mathrm{d}x$$

Now expand $$(1-\cos^2x)^{\frac{n-1}{2}}$$

You get a polynomial of form $$au'u^k$$, where $$u'=\sin(x)$$ and $$u^k=\cos^k$$ , which is integrable.

Well, my teacher gave us what is given below, without proof. If only this satisfies you, well and good, but if not, I'm gonna have to ask fellow members to help.

• Thank you, it is a reasonable answer for me. May 7, 2020 at 15:07
• I hope you understood the $1 or 2$ at the end of every bracketed term? May 7, 2020 at 15:13
• I know, it's okay. May 7, 2020 at 15:19