Question Number 144174 by lapache last updated on 22/Jun/21
$$\int_{\mathrm{0}} ^{\pi} \left({sinx}\right)^{\mathrm{2}{n}} {dx}=….?\:\:\:\forall{n}\in\mathbb{N} \\ $$
Commented by Willson last updated on 22/Jun/21
$$\mathrm{4}^{\mathrm{n}} \mathrm{sin}^{\mathrm{2n}} \mathrm{t}\:=\:\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} +\mathrm{2}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}−\mathrm{k}} \mathrm{cos}\left(\mathrm{2kt}\right) \\ $$$$\mathrm{4}^{\mathrm{n}} \underset{\:\mathrm{0}} {\int}^{\:\boldsymbol{\pi}} \mathrm{sin}^{\mathrm{2n}} \mathrm{t}\:\mathrm{dt}=\:\boldsymbol{\pi}\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \:+\mathrm{2}\underset{\mathrm{0}} {\underbrace{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \underset{\mathrm{0}} {\int}^{\:\boldsymbol{\pi}} \mathrm{cos}\left(\mathrm{2kt}\right)\mathrm{dt}}}\: \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\int}^{\:\pi} \left(\mathrm{sint}\right)^{\mathrm{2n}} \mathrm{dt}=\:\pi\left(\frac{\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} }{\mathrm{4}^{\mathrm{n}} }\right) \\ $$
Answered by Dwaipayan Shikari last updated on 22/Jun/21
$$\int_{\mathrm{0}} ^{\pi} \left({sinx}\right)^{\mathrm{2}{n}} {dx}=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \left({sinx}\right)^{\mathrm{2}{n}} {dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\mathrm{2}{n}} {dx} \\ $$$${x}={z}−\frac{\pi}{\mathrm{2}}\:\:\:\:\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(−{sin}\left(\frac{\pi}{\mathrm{2}}−{z}\right)\right)^{\mathrm{2}{n}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} \left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} \left({z}\right){dz}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} \left({x}\right){dx}\overset{{Dummy}\:{variable}} {=}\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} \left({x}\right){dx} \\ $$$${Now}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{a}−\mathrm{1}} \left({x}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left({x}\right){dx}=\frac{\Gamma\left({b}\right)\Gamma\left({a}\right)}{\mathrm{2}\Gamma\left({b}+{a}\right)} \\ $$$${Here}\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} \left({x}\right){dx}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}=\frac{\sqrt{\pi}}{{n}!}\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$