find-0-pi-t-2-sint-dt-

Question Number 27600 by abdo imad last updated on 10/Jan/18

f i n d \int_{0}^{π} \frac{t}{2 + s i n t} d t

Commented by abdo imad last updated on 11/Jan/18

l e t p u t I = \int_{0}^{π} \frac{t}{2 + s i n t} d t a n d t h e c h . t a n (\frac{t}{2}) = x \Leftrightarrow

t = 2 a r c t a n x

I = \int_{0}^{\infty} \frac{2 a r c t a n x}{2 + \frac{2 x}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}} = 2 \int_{0}^{\infty} \frac{a r c t a n x}{1 + x^{2} + 2 x} d x

= 2 \int_{0}^{\infty} \frac{a r c t a n x}{{(x + 1)}^{2}} d x t h e n t h e i n t e g r a t i o n p e r p a r t s g i v e

\int_{0}^{\infty} \frac{a r c t a n x}{{(x + 1)}^{2}} d x = {[- \frac{1}{x + 1} a r c t a n x]}_{0}^{\propto} - \int_{0}^{\infty} - \frac{1}{x + 1} \frac{d x}{1 + x^{2}}

= \int_{0}^{\infty} \frac{d x}{(x + 1) (1 + x^{2})} l e t d e c o m p o s e r a t i o n a l f r a c t i o n

F (x) = \frac{1}{(x + 1) (1 + x^{2})} \Rightarrow F (x) = \frac{a}{x + 1} + \frac{b x + c}{1 + x^{2}}

a = {l i m}_{x - > - 1^{}} (x + 1) F (x) = \frac{1}{2}

{l i m}_{x - >\propto} x F (x) = 0 = a + b \Rightarrow b = - \frac{1}{2} s o

F (x) = \frac{1}{2 (x + 1)} + \frac{\frac{- x}{2} + c}{1 + x^{2}}, F (0) = 1 = \frac{1}{2} + c \Rightarrow c = \frac{1}{2}

F (x) = \frac{1}{2 (x + 1)} - \frac{1}{2} \frac{x - 1}{1 + x^{2}}

I = \int_{0}^{\infty} \frac{d x}{x + 1} - \int_{0}^{\infty} \frac{x - 1}{1 + x^{2}} d x

= \int_{0}^{\infty} \frac{d x}{x + 1} - \frac{1}{2} \int_{0}^{\infty} \frac{2 x - 2}{1 + x^{2}} d x

I = \int_{0}^{\infty} (\frac{1}{x + 1} - \frac{1}{2} \frac{2 x}{1 + x^{2}}) d x + \int_{0}^{\infty} \frac{d x}{1 + x^{2}}

I = {[l n / x + 1 / - \frac{1}{2} l n / 1 + x^{2} /]}_{0}^{\propto} + \frac{π}{2}

I = {[l n / \frac{x + 1}{\sqrt{1 + x^{2}}} /]}_{0}^{\propto} + \frac{π}{2} = 0 + \frac{π}{2}

\int_{0}^{π} \frac{t}{2 + s i n t} d t = \frac{π}{2} .