find-the-value-of-0-pi-xdx-1-sinx-

Question Number 32349 by abdo imad last updated on 23/Mar/18

f i n d t h e v a l u e o f \int_{0}^{π} \frac{x d x}{1 + s i n x} .

Commented by abdo imad last updated on 24/Mar/18

t h e c h . t a n (\frac{x}{2}) = t g i v e

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

= 4 \int_{0}^{\infty} \frac{a r c t a n t}{1 + t^{2} + 2 t} d t = 4 \int_{0}^{\infty} \frac{a r c t a n t}{{(t + 1)}^{2}} d t^{'} b y p a r t s w e g e t

I = 4 ({[- \frac{1}{t + 1} a r c t a n t]}_{o}^{+ \infty} - \int_{0}^{\infty} - \frac{1}{t + 1} \frac{d t}{1 + t^{2}})

I = 4 \int_{0}^{\infty} \frac{d t}{(t + 1) (t^{2} + 1)} l e t d e c o m p o s e

f (t) = \frac{1}{(t + 1) (t^{2} + 1)} = \frac{a}{t + 1} + \frac{b t + c}{t^{2} + 1}

a = {l i m}_{t \to - 1} (t + 1) f (t) = \frac{1}{2}

{l i m}_{t \to + \infty} t f (t) = 0 = a + b \Rightarrow b = - a = - \frac{1}{2} \Rightarrow

f (t) = \frac{1}{2 (t + 1)} + \frac{\frac{- 1}{2} t + c}{t^{2} + 1}

f (0) = 1 = \frac{1}{2} + c \Rightarrow c = \frac{1}{2} \Rightarrow f (t) = \frac{1}{2 (t + 1)} + \frac{1}{2} \frac{- t + 1}{t^{2} + 1}

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

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

I = 2 {[l n ∣ t + 1 ∣ - l n \sqrt{t^{2} + 1}]}_{0}^{\infty} + π

I = 2 {[l n ∣ \frac{t + 1}{\sqrt{t^{2} + 1}} ∣]}_{0}^{\infty} + π = 0 + π \Rightarrow I = π .