∫_0 ^π ((x^2 cos x)/((1+sin x)^2 )) dx ?


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Question Number 104220 by bemath last updated on 20/Jul/20

$\underset{0}{\int^{π}} \frac{x^{2} \cos x}{{(1 + \sin x)}^{2}} d x ?$
Commented by bemath last updated on 20/Jul/20

$t h a n k y o u b o t h . c o o l l$
	Answered by Ar Brandon last updated on 20/Jul/20
	$I=∫_0 ^π ((x^2 cosx)/((1+sinx)^2 ))dx , x=π−u I=∫_0 ^π (((π−x)^2 cos(π−x))/((1+sin(π−x))^2 ))dx=−∫(((π−x)^2 cosx)/((1+sinx)^2 ))dx 2I=∫_0 ^π ((x^2 −(π^2 −2πx+x^2 ))/((1+sinx)^2 ))∙cosxdx=∫_0 ^π ((2πx−π^2 )/((1+sinx)^2 ))cosxdx =∫_0 ^π {((2πxcosx)/((1+sinx)^2 ))−((π^2 cosx)/((1+sinx)^2 ))}dx=∫_0 ^π ((2πxcosx)/((1+sinx)^2 ))dx+[(π^2 /(1+sinx))]_0 ^π u(x)=2πx⇒u′(x)=2π , v′(x)=((cosx)/((1+sinx)^2 ))⇒v(x)=((−1)/(1+sinx)) ⇒2I={((−2πx)/(1+sinx))+∫((2πdx)/(1+sinx))}_0 ^π ∫(dx/(1+sinx))=∫((1−sinx)/(cos^2 x))dx=tanx−secx ⇒2I=2π{((−x)/(1+sinx))+tanx−secx}_0 ^π =2π(2−π) ⇒∫_0 ^π ((x^2 cosx)/((1+sinx)^2 ))dx=(2−π)π$
	$I = \int_{0}^{π} \frac{x^{2} cosx}{{(1 + sinx)}^{2}} dx, x = π - u$ $I = \int_{0}^{π} \frac{{(π - x)}^{2} \cos (π - x)}{{(1 + \sin (π - x))}^{2}} dx = - \int \frac{{(π - x)}^{2} cosx}{{(1 + sinx)}^{2}} dx$ $2 I = \int_{0}^{π} \frac{x^{2} - (π^{2} - 2 π x + x^{2})}{{(1 + sinx)}^{2}} \cdot cosxdx = \int_{0}^{π} \frac{2 π x - π^{2}}{{(1 + sinx)}^{2}} cosxdx$ $= \int_{0}^{π} {\frac{2 π xcosx}{{(1 + sinx)}^{2}} - \frac{π^{2} cosx}{{(1 + sinx)}^{2}}} dx = \int_{0}^{π} \frac{2 π xcosx}{{(1 + sinx)}^{2}} dx + {[\frac{π^{2}}{1 + sinx}]}_{0}^{π}$ $u (x) = 2 π x \Rightarrow u^{'} (x) = 2 π, v^{'} (x) = \frac{cosx}{{(1 + sinx)}^{2}} \Rightarrow v (x) = \frac{- 1}{1 + sinx}$ $\Rightarrow 2 I = {\frac{- 2 π x}{1 + sinx} + \int \frac{2 π dx}{1 + sinx}}_{0}^{π}$ $\int \frac{dx}{1 + sinx} = \int \frac{1 - sinx}{\cos^{2} x} dx = tanx - secx$ $\Rightarrow 2 I = 2 π {\frac{- x}{1 + sinx} + tanx - secx}_{0}^{π} = 2 π (2 - π)$ $\Rightarrow \int_{0}^{π} \frac{x^{2} cosx}{{(1 + sinx)}^{2}} dx = (2 - π) π$
	Answered by john santu last updated on 20/Jul/20
	$by parts { ((u=x^2 ⇒du=2x dx)),((v=∫ ((d(1+sin x))/((1+sin x)^2 )) =−(1/(1+sin x)) )) :} I= −(x^2 /(1+sin x)) ]_0 ^π +∫_0 ^π ((2x)/(1+sin x)) dx I= −π^2 +∫_0 ^π ((2x)/(1+sin x)) dx set J = ∫_0 ^π ((2x)/(1+sin x)) dx replace x by π−x J = ∫_π ^0 ((2(π−x))/(1+sin (π−x))) (−dx) J=∫_0 ^π ((2π−2x)/(1+sin x)) dx , so we have 2J = ∫_0 ^π ((2π)/(1+sin x)) dx ⇒J = ∫_0 ^π (π/(1+sin x))dx substitute t=x−(π/2) J=∫_0 ^(π/2) π sec^2 ((t/2)) dt = 2π hence we conclude that = −π^2 +2π (JS ⊛)$
	$b y p a r t s$ ${\begin{cases} u = x^{2} \Rightarrow d u = 2 x d x \\ v = \int \frac{d (1 + \sin x)}{{(1 + \sin x)}^{2}} = - \frac{1}{1 + \sin x} \end{cases}$ ${I = - \frac{x^{2}}{1 + \sin x}]}_{0}^{π} + \underset{0}{\int^{π}} \frac{2 x}{1 + \sin x} d x$ $I = - π^{2} + \underset{0}{\int^{π}} \frac{2 x}{1 + \sin x} d x$ $s e t J = \underset{0}{\int^{π}} \frac{2 x}{1 + \sin x} d x$ $r e p l a c e x b y π - x$ $J = \underset{π}{\int^{0}} \frac{2 (π - x)}{1 + \sin (π - x)} (- d x)$ $J = \underset{0}{\int^{π}} \frac{2 π - 2 x}{1 + \sin x} d x, s o w e h a v e$ $2 J = \underset{0}{\int^{π}} \frac{2 π}{1 + \sin x} d x \Rightarrow J = \underset{0}{\int^{π}} \frac{π}{1 + \sin x} d x$ $s u b s t i t u t e t = x - \frac{π}{2}$ $J = \underset{0}{\int^{π / 2}} π \sec^{2} (\frac{t}{2}) d t = 2 π$ $h e n c e w e c o n c l u d e t h a t$ $= - π^{2} + 2 π (J S ⊛)$
	Answered by mathmax by abdo last updated on 20/Jul/20
	$I =∫_0 ^π ((x^2 cosx)/((1+sinx)^2 ))dx changement x =π−t give I =−∫_0 ^π (((π−t)^2 (cos(π−t)))/((1+sint)^2 ))(−dt) =−∫_0 ^π (((π^2 −2πt +t^2 )cost)/((1+sint)^2 ))dt =−π^2 ∫_0 ^π ((cost)/((1+sint)^2 )) +2π ∫_0 ^π ((tcost)/((1+sint)^2 ))dt −∫_0 ^π ((t^2 cost)/((1+sint)^2 ))dt ⇒ 2I =π^2 [(1/(1+sint))]_0 ^π +2π ∫_0 ^π ((tcost)/((1+sint)^2 ))dt =0 +2π ∫_0 ^(π ) ((tcost)/((1+sint)^2 ))dx by parts u^′ =((cost)/((1+sint)^2 )) and v =t ⇒ ∫_0 ^π ((tcost)/((1+sint)^2 ))dt =[−(t/(1+sint))]_0 ^π −∫_0 ^π −(1/(1+sint))dt =−π +∫_0 ^π (dt/(1+sint)) changement tan((t/2))=u give ∫_0 ^π (dt/(1+sint)) =∫_0 ^∞ ((2du)/((1+u^2 )(1+((2u)/(1+u^2 ))))) =∫_0 ^∞ ((2du)/(1+u^2 +2u)) =∫_0 ^∞ ((2du)/((u+1)^2 )) =[((−2)/(u+1))]_0 ^∞ =2 ⇒2I =2π{−π +2} ⇒I =2π−π^2$
	$I = \int_{0}^{π} \frac{x^{2} cosx}{{(1 + sinx)}^{2}} dx changement x = π - t give$ $I = - \int_{0}^{π} \frac{{(π - t)}^{2} (\cos (π - t))}{{(1 + sint)}^{2}} (- dt) = - \int_{0}^{π} \frac{(π^{2} - 2 π t + t^{2}) cost}{{(1 + sint)}^{2}} dt$ $= - π^{2} \int_{0}^{π} \frac{cost}{{(1 + sint)}^{2}} + 2 π \int_{0}^{π} \frac{tcost}{{(1 + sint)}^{2}} dt - \int_{0}^{π} \frac{t^{2} cost}{{(1 + sint)}^{2}} dt \Rightarrow$ $2 I = π^{2} {[\frac{1}{1 + sint}]}_{0}^{π} + 2 π \int_{0}^{π} \frac{tcost}{{(1 + sint)}^{2}} dt$ $= 0 + 2 π \int_{0}^{π} \frac{tcost}{{(1 + sint)}^{2}} dx by parts u^{^{'}} = \frac{cost}{{(1 + sint)}^{2}} and v = t \Rightarrow$ $\int_{0}^{π} \frac{tcost}{{(1 + sint)}^{2}} dt = {[- \frac{t}{1 + sint}]}_{0}^{π} - \int_{0}^{π} - \frac{1}{1 + sint} dt$ $= - π + \int_{0}^{π} \frac{dt}{1 + sint} changement \tan (\frac{t}{2}) = u give$ $\int_{0}^{π} \frac{dt}{1 + sint} = \int_{0}^{\infty} \frac{2 du}{(1 + u^{2}) (1 + \frac{2 u}{1 + u^{2}})} = \int_{0}^{\infty} \frac{2 du}{1 + u^{2} + 2 u} = \int_{0}^{\infty} \frac{2 du}{{(u + 1)}^{2}} = {[\frac{- 2}{u + 1}]}_{0}^{\infty} = 2$ $\Rightarrow 2 I = 2 π {- π + 2} \Rightarrow I = 2 π - π^{2}$
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