find-0-pi-xdx-1-sinx-

Question Number 35242 by abdo mathsup 649 cc last updated on 17/May/18

f i n d \int_{0}^{π} \frac{x d x}{1 + s i n x}

Commented by abdo mathsup 649 cc last updated on 17/May/18

l e t p u t I = \int_{0}^{π} \frac{x d x}{1 + s i n x} . c h a n g e m e n t t a n (\frac{x}{2}) = t

g i v e I = \int_{0}^{\infty} \frac{2 a r c 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

a n d b y p a r t s

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

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

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

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

= {[l n (\frac{t + 1}{t^{2} + 1})]}_{0}^{+ \infty} + \frac{π}{2} = \frac{π}{2} \Rightarrow I = 4 . \frac{π}{2} \Rightarrow

I = 2 π .

Answered by math1967 last updated on 17/May/18

2 I = \underset{0}{\int^{π}} \frac{x d x}{1 + s i n x} + \underset{0}{\int^{π}} \frac{(π - x) d x}{1 + s i n (π - x)}

= π \underset{0}{\int^{π}} \frac{d x}{1 + s i n x} = π \underset{0}{\int^{π}} \frac{d x}{1 + c o s (\frac{π}{2} - x)}

= π \underset{0}{\int^{π}} \frac{d x}{2 {c o s}^{2} (\frac{π}{4} - \frac{x}{2})} [1 + c o s x = 2 {c o s}^{2} \frac{x}{2}]

= π \underset{0}{\int^{π}} {s e c}^{2} (\frac{π}{4} - \frac{x}{2}) d x

= \frac{π}{2} \underset{0}{\int^{π}} {s e c}^{2} (\frac{π}{4} - \frac{x}{2}) d x = \frac{π}{2} {[\frac{t a n (\frac{π}{4} - \frac{x}{2})}{\frac{- 1}{2}}]}_{0}^{π}

= - π {t a n (- \frac{π}{4}) - t a n \frac{π}{4}} = 2 π

∴ I = π

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