calculate-0-pi-tsint-3-sin-2-t-dt-

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

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}