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$$\int_0^{\frac{\pi}{2}}\cos^{n}(t)dt =?$$ 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.

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3 Answers 3

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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 $$I(n)=\frac{n-1}{n}\int_0^{\frac{\pi}{2}}\cos^{n-2}(t)dt=\frac{n-1}{n}I(n-2)$$

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  • $\begingroup$ Thank you, In this way I will solve the question. $\endgroup$ May 7, 2020 at 15:09
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$$\cos^n(t)=\sin^n\left(\frac{\pi}{2}-t\right)$$

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.

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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.enter image description here

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

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