BeMath-I-0-pi-x-dx-1-sin-x-

Question Number 108391 by bemath last updated on 16/Aug/20

\frac{∡ B e M a t h ∡}{▽}

I = \underset{0}{\int^{π}} \frac{x d x}{1 + \sin x}

Answered by bobhans last updated on 16/Aug/20

Commented by khanshadab2405 last updated on 16/Aug/20

1/√1²+cot²∅

Commented by bobhans last updated on 16/Aug/20

t y p o I = \frac{1}{2} \underset{- π / 4}{\int^{π / 4}} (4 w + π) \sec^{2} w d w

I = \frac{1}{2} \times 2 π = π

Answered by Dwaipayan Shikari last updated on 16/Aug/20

\int_{0}^{π} \frac{x d x}{1 + s i n x} = \int_{0}^{π} \frac{π - x}{1 + s i n x} d x = I

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

2 I = 2 π \int_{0}^{\infty} \frac{1}{1 + \frac{2 t}{1 + t^{2}}} . \frac{1}{1 + t^{2}} d t (t a n \frac{x}{2} = t

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

I = π {[- \frac{1}{1 + t}]}_{0}^{\infty}

I = π

Answered by mnjuly1970 last updated on 16/Aug/20

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

\Rightarrow 2 I = π \int_{0}^{π} \frac{1 - s i n (x)}{{c o s}^{2} (x)} d x

2 I = π ({[t a n (x)]}_{0}^{π} - {[\frac{1}{c o s (x)}]}_{0}^{π}) = π (2)

I = π ♣ \dots \dots \dots . . ♣

Answered by mathmax by abdo last updated on 16/Aug/20

I = \int_{0}^{π} \frac{xdx}{1 + sinx} changement x = π - t give

I = \int_{0}^{π} \frac{(π - t)}{1 + sint} dt = π \int_{0}^{π} \frac{dt}{1 + sint} - I \Rightarrow 2 I = π \int_{0}^{π} \frac{dt}{1 + sint}

=_{\tan (\frac{t}{2}) = u} π \int_{0}^{\infty} \frac{2 du}{(1 + u^{2}) (1 + \frac{2 u}{1 + u^{2}})}

= 2 π \int_{0}^{\infty} \frac{du}{1 + u^{2} + 2 u} = 2 π \int_{0}^{\infty} \frac{du}{{(u + 1)}^{2}} = 2 π {[- \frac{1}{u + 1}]}_{0}^{+ \infty} = 2 π \Rightarrow

2 I = 2 π \Rightarrow I = π

Answered by mathmax by abdo last updated on 16/Aug/20

another way changement \tan (\frac{x}{2}) = t give

I = \int_{0}^{\infty} \frac{2 \arctan (t)}{(1 + \frac{2 t}{1 + t^{2}})} \frac{2 dt}{1 + t^{2}} = 4 \int_{0}^{\infty} \frac{arctant}{1 + t^{2} + 2 t} dt

= 4 \int_{0}^{\infty} \frac{arctant}{{(t + 1)}^{2}} dt =_{byparts} 4 {{[- \frac{arctant}{t + 1}]}_{0}^{+ \infty} + \int_{0}^{\infty} \frac{1}{t + 1} \frac{dt}{1 + t^{2}}}

= 4 \int_{0}^{\infty} \frac{dt}{(t + 1) (t^{2} + 1)} let decompose F (t) = \frac{1}{(t + 1) (t^{2} + 1)}

F (t) = \frac{a}{t + 1} + \frac{bt + c}{t^{2} + 1}

a = \frac{1}{2}, \lim_{t \to + \infty} tF (t) = 0 = a + b \Rightarrow b = - \frac{1}{2}

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

\int_{0}^{\infty} F (t) dt = \frac{1}{2} \int_{0}^{\infty} \frac{dt}{t + 1} - \frac{1}{4} \int_{0}^{\infty} \frac{2 t}{t^{2} + 1} + \frac{1}{2} \int_{0}^{\infty} \frac{dt}{1 + t^{2}}

= {[\frac{1}{2} \ln (t + 1) - \frac{1}{4} \ln (t^{2} + 1)]}_{0}^{\infty} + \frac{1}{2} {[arctant]}_{0}^{\infty}

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