calculate ∫_0 ^π ((x^2 cosx)/(3+sin^2 x))dx


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Question Number 76779 by mathmax by abdo last updated on 30/Dec/19

$c a l c u l a t e \int_{0}^{π} \frac{x^{2} c o s x}{3 + {s i n}^{2} x} d x$
Commented by mathmax by abdo last updated on 10/Jan/20
$let A =∫_0 ^π ((x^2 cosx)/(3+sin^2 x))dx changement tan((x/2))=t give A =∫_0 ^∞ ((4 arctan^2 (t)×((1−t^2 )/(1+t^2 )))/(3+((4t^2 )/((1+t^2 )^2 ))))×((2dt)/(1+t^2 )) =8 ∫_0 ^∞ (((1−t^2 )arctan^2 t)/((1+t^2 )^2 (3+((4t^2 )/((1+t^2 )^2 )))))dt =8 ∫_0 ^∞ (((1−t^2 )arctan^2 t)/(3(t^2 +1)^2 +4t^2 ))dt =8 ∫_0 ^∞ (((1−t^2 )(arctant)^2 )/(3(t^4 +2t^2 +1)+4t^2 )) =8 ∫_0 ^∞ (((1−t^2 )(arctant)^2 )/(3t^4 +10t^2 +3))dt let solve 3t^4 +10t^2 +3=0 ⇒3u^4 +10u +3=0 →Δ^′ =5^2 −9 =16 ⇒u_1 =((−5+4)/3) =−(1/3) u_2 =((−5−4)/3) =−3 ⇒3t^4 +10t^2 +3 =3(t^2 +(1/3))(t^2 +3) ⇒ A =8 ∫_0 ^∞ (((1−t^2 )(arctant)^2 )/(3(t^2 +(1/3))(t^2 +3)))dt ⇒ (3/4) A =∫_(−∞) ^∞ (((1−t^2 )(arctant)^2 )/((t^2 +(1/3))(t^2 +3)))dt letW(z)=(((1−z^2 )(arctanz)^2 )/((z^2 +(1/3))(z^2 +3))) ⇒ W(z) =(((1−z^2 )(arctanz)^2 )/((z−(i/(√3)))(z+(i/(√3)))(z−i(√3))(z+i(√3)))) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,(i/(√3)))+Res(W,i(√3))} ...be continued...$
$l e t A = \int_{0}^{π} \frac{x^{2} c o s x}{3 + {s i n}^{2} x} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $A = \int_{0}^{\infty} \frac{4 {a r c t a n}^{2} (t) \times \frac{1 - t^{2}}{1 + t^{2}}}{3 + \frac{4 t^{2}}{{(1 + t^{2})}^{2}}} \times \frac{2 d t}{1 + t^{2}} = 8 \int_{0}^{\infty} \frac{(1 - t^{2}) {a r c t a n}^{2} t}{{(1 + t^{2})}^{2} (3 + \frac{4 t^{2}}{{(1 + t^{2})}^{2}})} d t$ $= 8 \int_{0}^{\infty} \frac{(1 - t^{2}) {a r c t a n}^{2} t}{3 {(t^{2} + 1)}^{2} + 4 t^{2}} d t = 8 \int_{0}^{\infty} \frac{(1 - t^{2}) {(a r c t a n t)}^{2}}{3 (t^{4} + 2 t^{2} + 1) + 4 t^{2}}$ $= 8 \int_{0}^{\infty} \frac{(1 - t^{2}) {(a r c t a n t)}^{2}}{3 t^{4} + 10 t^{2} + 3} d t l e t s o l v e 3 t^{4} + 10 t^{2} + 3 = 0$ $\Rightarrow 3 u^{4} + 10 u + 3 = 0 \to Δ^{^{'}} = 5^{2} - 9 = 16 \Rightarrow u_{1} = \frac{- 5 + 4}{3} = - \frac{1}{3}$ $u_{2} = \frac{- 5 - 4}{3} = - 3 \Rightarrow 3 t^{4} + 10 t^{2} + 3 = 3 (t^{2} + \frac{1}{3}) (t^{2} + 3) \Rightarrow$ $A = 8 \int_{0}^{\infty} \frac{(1 - t^{2}) {(a r c t a n t)}^{2}}{3 (t^{2} + \frac{1}{3}) (t^{2} + 3)} d t \Rightarrow$ $\frac{3}{4} A = \int_{- \infty}^{\infty} \frac{(1 - t^{2}) {(a r c t a n t)}^{2}}{(t^{2} + \frac{1}{3}) (t^{2} + 3)} d t l e t W (z) = \frac{(1 - z^{2}) {(a r c t a n z)}^{2}}{(z^{2} + \frac{1}{3}) (z^{2} + 3)} \Rightarrow$ $W (z) = \frac{(1 - z^{2}) {(a r c t a n z)}^{2}}{(z - \frac{i}{\sqrt{3}}) (z + \frac{i}{\sqrt{3}}) (z - i \sqrt{3}) (z + i \sqrt{3})} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, \frac{i}{\sqrt{3}}) + R e s (W, i \sqrt{3})}$ $. . . b e c o n t i n u e d . . .$
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