calculate ∫_0 ^π ((tsint)/(3+sin^2 t)) dt


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Question Number 64068 by mathmax by abdo last updated on 12/Jul/19

$c a l c u l a t e \int_{0}^{π} \frac{t s i n t}{3 + {s i n}^{2} t} d t$
Commented by mathmax by abdo last updated on 17/Jul/19
$let A =∫_0 ^π ((tsint)/(3 +sin^2 t))dt changement t =π−x give A =−∫_0 ^π (((π−x)sin(π−x))/(3+sin^2 (π−x))) (−dx) =∫_0 ^π ((πsinx)/(3+sin^2 x))dx−∫_0 ^π ((xsinx)/(3+sin^2 x))dx ⇒2A =π ∫_0 ^π ((sinx)/(3+sin^2 x)) dx ⇒A =(π/2) ∫_0 ^π ((sinx )/(3+sin^2 x))dx changement tan((x/2)) =u give ∫_0 ^π ((sinx)/(3+sin^2 x))dx =∫_0 ^∞ (((2u)/(1+u^2 ))/(3 +((4u^2 )/((1+u^2 )^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^∞ ((4u)/((1+u^2 )^2 {3 +((4u^2 )/((1+u^2 )^2 ))})) =∫_0 ^∞ ((4u)/(3(1+u^2 )^2 +4u^2 ))du =∫_0 ^∞ ((4u)/(3(u^4 +2u^2 +1) +4u^2 ))du =∫_0 ^∞ ((4u)/(3u^4 +6u^2 +3 +4u^2 ))du =∫_0 ^∞ ((4u)/(3u^4 +10u^2 +3))du 3u^4 +10 u^2 +3 =0 ⇒3x^2 +10x +3 =0 (x=u^2 ) Δ^′ =25−9 =16 ⇒x_1 =((−5 +4)/3) =−(1/3) x_2 =((−5−4)/3) =−3 ⇒3x^2 +10x +3 =3(x+(1/3))(x+3) ⇒ 3u^4 +10u^2 +3 =3(u^2 +(1/3))(u^2 +3) ⇒ ((4u)/(3u^4 +10u^2 +3)) =((4u)/3)(3/8){(1/(u^2 +(1/3)))−(1/(u^2 +3))} =(u/2){ (1/(u^2 +(1/3)))−(1/(u^2 +3))} ⇒ ∫_0 ^∞ ((4u)/(3u^4 +10u^2 +3))du =(1/2){ ∫_0 ^∞ ((udu)/(u^2 +(1/3))) −∫_0 ^∞ ((u du)/(u^2 +3))} =(1/4)[ln∣((u^2 +(1/3))/(u^2 +3))∣]_0 ^(+∞) =(1/4)(−ln((1/9)))=((ln9)/4) =((2ln3)/4) =((ln3)/2) ⇒ A =(π/2) ((ln3)/2) ⇒ A =((πln3)/4)$
$l e t A = \int_{0}^{π} \frac{t s i n t}{3 + {s i n}^{2} t} d t c h a n g e m e n t t = π - x g i v e$ $A = - \int_{0}^{π} \frac{(π - x) s i n (π - x)}{3 + {s i n}^{2} (π - x)} (- d x) = \int_{0}^{π} \frac{π s i n x}{3 + {s i n}^{2} x} d x - \int_{0}^{π} \frac{x s i n x}{3 + {s i n}^{2} x} d x$ $\Rightarrow 2 A = π \int_{0}^{π} \frac{s i n x}{3 + {s i n}^{2} x} d x \Rightarrow A = \frac{π}{2} \int_{0}^{π} \frac{s i n x}{3 + {s i n}^{2} x} d x$ $c h a n g e m e n t t a n (\frac{x}{2}) = u g i v e$ $\int_{0}^{π} \frac{s i n x}{3 + {s i n}^{2} x} d x = \int_{0}^{\infty} \frac{\frac{2 u}{1 + u^{2}}}{3 + \frac{4 u^{2}}{{(1 + u^{2})}^{2}}} \frac{2 d u}{1 + u^{2}}$ $= \int_{0}^{\infty} \frac{4 u}{{(1 + u^{2})}^{2} {3 + \frac{4 u^{2}}{{(1 + u^{2})}^{2}}}} = \int_{0}^{\infty} \frac{4 u}{3 {(1 + u^{2})}^{2} + 4 u^{2}} d u$ $= \int_{0}^{\infty} \frac{4 u}{3 (u^{4} + 2 u^{2} + 1) + 4 u^{2}} d u = \int_{0}^{\infty} \frac{4 u}{3 u^{4} + 6 u^{2} + 3 + 4 u^{2}} d u$ $= \int_{0}^{\infty} \frac{4 u}{3 u^{4} + 10 u^{2} + 3} d u$ $3 u^{4} + 10 u^{2} + 3 = 0 \Rightarrow 3 x^{2} + 10 x + 3 = 0 (x = u^{2})$ $Δ^{^{'}} = 25 - 9 = 16 \Rightarrow x_{1} = \frac{- 5 + 4}{3} = - \frac{1}{3}$ $x_{2} = \frac{- 5 - 4}{3} = - 3 \Rightarrow 3 x^{2} + 10 x + 3 = 3 (x + \frac{1}{3}) (x + 3) \Rightarrow$ $3 u^{4} + 10 u^{2} + 3 = 3 (u^{2} + \frac{1}{3}) (u^{2} + 3) \Rightarrow$ $\frac{4 u}{3 u^{4} + 10 u^{2} + 3} = \frac{4 u}{3} \frac{3}{8} {\frac{1}{u^{2} + \frac{1}{3}} - \frac{1}{u^{2} + 3}} = \frac{u}{2} {\frac{1}{u^{2} + \frac{1}{3}} - \frac{1}{u^{2} + 3}} \Rightarrow$ $\int_{0}^{\infty} \frac{4 u}{3 u^{4} + 10 u^{2} + 3} d u = \frac{1}{2} {\int_{0}^{\infty} \frac{u d u}{u^{2} + \frac{1}{3}} - \int_{0}^{\infty} \frac{u d u}{u^{2} + 3}}$ $= \frac{1}{4} {[l n ∣ \frac{u^{2} + \frac{1}{3}}{u^{2} + 3} ∣]}_{0}^{+ \infty} = \frac{1}{4} (- l n (\frac{1}{9})) = \frac{l n 9}{4} = \frac{2 l n 3}{4} = \frac{l n 3}{2} \Rightarrow$ $A = \frac{π}{2} \frac{l n 3}{2} \Rightarrow A = \frac{π l n 3}{4}$
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