∫_0^ ^(Π/2) xcos^n xdx by reduction formula


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Question Number 74037 by akshaypalsra8@gmail.com last updated on 18/Nov/19

$\int_{0^{}}^{Π / 2} x \cos^{n} x d x b y r e d u c t i o n f o r m u l a$
	Answered by mind is power last updated on 18/Nov/19
	$I_n =∫_0 ^(π/2) xcos^n (x)dx ⇒I_n −I_(n+2) =∫_0 ^(π/2) xcos^n (x).sin^2 (x)dx by part u(x)=xsin(x),v′(x)=cos^n (x)sin(x) I_n −I_(n+2) =[((xsin(x).cos^(n+1) (x))/(−(n+1)))]_0 ^(π/2) +(1/(n+1))∫_0 ^(π/2) cos^(n+1) (x).{sin(x)+xcos(x)}dx =(1/(n+1))∫_0 ^(π/2) cos^(n+1) (x)sin(x)dx+(1/(n+1))∫_0 ^(π/2) xcos^(n+2) (x)dx =(1/(n+1)).[((−cos^(n+2) (x))/(n+2))]_0 ^(π/2) +(I_(n+2) /(n+2))=I_n −I_(n+2) ⇒(1/((n+1)(n+2)))+(I_(n+2) /(n+2))=I_n −I_(n+2) ⇔I_(n+2) (n+3)(n+1)−(n+1)(n+2)I_n =1$
	$I_{n} = \int_{0}^{\frac{π}{2}} {xcos}^{n} (x) dx$ $\Rightarrow I_{n} - I_{n + 2} = \int_{0}^{\frac{π}{2}} {xcos}^{n} (x) . \sin^{2} (x) dx$ $by part$ $u (x) = xsin (x), v^{'} (x) = \cos^{n} (x) \sin (x)$ $I_{n} - I_{n + 2} = {[\frac{xsin (x) . \cos^{n + 1} (x)}{- (n + 1)}]}_{0}^{\frac{π}{2}} + \frac{1}{n + 1} \int_{0}^{\frac{π}{2}} \cos^{n + 1} (x) . {\sin (x) + xcos (x)} dx$ $= \frac{1}{n + 1} \int_{0}^{\frac{π}{2}} \cos^{n + 1} (x) \sin (x) dx + \frac{1}{n + 1} \int_{0}^{\frac{π}{2}} {xcos}^{n + 2} (x) dx$ $= \frac{1}{n + 1} . {[\frac{- \cos^{n + 2} (x)}{n + 2}]}_{0}^{\frac{π}{2}} + \frac{I_{n + 2}}{n + 2} = I_{n} - I_{n + 2}$ $\Rightarrow \frac{1}{(n + 1) (n + 2)} + \frac{I_{n + 2}}{n + 2} = I_{n} - I_{n + 2}$ $\Leftrightarrow I_{n + 2} (n + 3) (n + 1) - (n + 1) (n + 2) I_{n} = 1$
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