0-pi-2-xcos-x-1-sin-x-2-dx-

Question Number 80416 by jagoll last updated on 03/Feb/20

\underset{0}{\int^{\frac{π}{2}}} \frac{x \cos x}{{(1 + \sin x)}^{2}} d x ?

Answered by MJS last updated on 03/Feb/20

\int \frac{x \cos x}{{(1 + \sin x)}^{2}} d x =

by parts

u = x \to u^{'} = 1

v^{'} = \frac{\cos x}{{(1 + \sin x)}^{2}} \to v = - \frac{1}{1 + \sin x}

= - \frac{x}{1 + \sin x} + \int \frac{d x}{1 + \sin x} =

= - \frac{x}{1 + \sin x} + \tan x - \frac{1}{\cos x} + C

\underset{0}{\int^{\frac{π}{2}}} \frac{x \cos x}{{(1 + \sin x)}^{2}} d x =

[\lim_{x \to \frac{π}{2}} (\tan x - \frac{1}{\cos x}) = 0]

= 1 - \frac{π}{4}

Commented by jagoll last updated on 03/Feb/20

\frac{- x}{1 + \sin x} ∣_{0}^{\frac{π}{2}} = \frac{- \frac{π}{2}}{2} = - \frac{π}{4} .

h o w g e t 1 - \frac{π}{4} s i r ?

Commented by jagoll last updated on 03/Feb/20

o o i u n d e r s t a n d s i r . g e t 1

Facebook Tweet Pin