Question Number 15948 by tawa tawa last updated on 15/Jun/17
Commented by rahul 19 last updated on 13/Nov/18
Nice Question!
will anybody try?
I think this Q. can be done by
integration by parts...
Answered by Smail last updated on 13/Nov/18
![I_n =∫_0 ^(π/2) cos^(2n) xdx=∫_0 ^(π/2) cos^2 xcos^(2n−2) dx =∫_0 ^(π/2) cos^(2n−2) xdx−∫_0 ^(π/2) sin^2 xcos^(2n−2) xdx =I_(n−1) −∫sin(x)(sin(x)cos^(2n−2) (x)dx by parts u=sinx⇒u′=cosx v′=sin(x)cos^(2n−2) x⇒v=−(1/(2n−1))cos^(2n−1) I_n =I_(n−1) +[(1/(2n−1))sin(x)cos^(2n−1) x]_0 ^(π/2) −(1/(2n−1))∫_0 ^(π/2) cos^(2n) dx =I_(n−1) −(1/(2n−1))I_n I_n (1+(1/(2n−1)))=I_(n−1) ⇔I_n =((2n−1)/(2n))I_(n−1) I_(n−1) =((2n−3)/(2n−2))I_(n−2) I_n =(((2n−1)(2n−3))/(2^2 n(n−1)))I_(n−2) Generally I_n =(((2n−1)(2n−3)(2n−5)...1)/(2^n n!))I_0 =((2n(2n−1)(2n−2)(2n−3)(2n−4)(2n−5)...1)/(2^n n!(2n(2n−2)(2n−4)...1))((π/2)) =(((2n)!)/(2^n (n!)(2^n n(n−1)(n−2)...)))((π/2)) I_n =((π(2n)!)/(2^(2n+1) (n!)^2 ))](https://www.tinkutara.com/question/Q47716.png)
$${I}_{{n}} =\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{2}{n}} {xdx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{2}} {xcos}^{\mathrm{2}{n}−\mathrm{2}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{2}{n}−\mathrm{2}} {xdx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}^{\mathrm{2}} {xcos}^{\mathrm{2}{n}−\mathrm{2}} {xdx} \\ $$$$={I}_{{n}−\mathrm{1}} −\int{sin}\left({x}\right)\left({sin}\left({x}\right){cos}^{\mathrm{2}{n}−\mathrm{2}} \left({x}\right){dx}\right. \\ $$$${by}\:{parts} \\ $$$${u}={sinx}\Rightarrow{u}'={cosx} \\ $$$${v}'={sin}\left({x}\right){cos}^{\mathrm{2}{n}−\mathrm{2}} {x}\Rightarrow{v}=−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}{cos}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$${I}_{{n}} ={I}_{{n}−\mathrm{1}} +\left[\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}{sin}\left({x}\right){cos}^{\mathrm{2}{n}−\mathrm{1}} {x}\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{2}{n}} {dx} \\ $$$$={I}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}{I}_{{n}} \\ $$$${I}_{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\right)={I}_{{n}−\mathrm{1}} \Leftrightarrow{I}_{{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}{I}_{{n}−\mathrm{1}} \\ $$$${I}_{{n}−\mathrm{1}} =\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{I}_{{n}−\mathrm{2}} \\ $$$${I}_{{n}} =\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{3}\right)}{\mathrm{2}^{\mathrm{2}} {n}\left({n}−\mathrm{1}\right)}{I}_{{n}−\mathrm{2}} \\ $$$${Generally} \\ $$$${I}_{{n}} =\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{3}\right)\left(\mathrm{2}{n}−\mathrm{5}\right)…\mathrm{1}}{\mathrm{2}^{{n}} {n}!}{I}_{\mathrm{0}} \\ $$$$=\frac{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)\left(\mathrm{2}{n}−\mathrm{3}\right)\left(\mathrm{2}{n}−\mathrm{4}\right)\left(\mathrm{2}{n}−\mathrm{5}\right)…\mathrm{1}}{\mathrm{2}^{{n}} {n}!\left(\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{2}\right)\left(\mathrm{2}{n}−\mathrm{4}\right)…\mathrm{1}\right.}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} \left({n}!\right)\left(\mathrm{2}^{{n}} {n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\right)}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$${I}_{{n}} =\frac{\pi\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \left({n}!\right)^{\mathrm{2}} } \\ $$$$ \\ $$