Mathematics

Solve: $$I=\int _{ o }^{ \pi  }{ \dfrac { xdx }{ 1+\sin { x }  }  }$$


SOLUTION
$$I=\displaystyle \int_0^n \dfrac {x\ dx}{1+\sin^2x}$$
$$=\displaystyle \int _0^n \dfrac {x}{\cos^2x / (1-\sin x)}$$
$$=\displaystyle \int \dfrac {x}{\cos^2 x}-\dfrac {x \sin x}{\cos ^2 s}dx$$
$$I_1=2.=\displaystyle \int \dfrac {x}{1+\cos 2x}$$
$$2=\displaystyle \int \dfrac {1}{2} x\tan x-\displaystyle \int \dfrac {1}{2} \tan \ dx$$
$$x\tan x- (-\ln |\cos x|)$$
$$I_2 =\displaystyle \int \dfrac {x \sin x}{\cos ^2 x} dx$$
$$=x \sin x \tan x-\displaystyle \int (\sin x+x \cos x)\tan x\ dx$$
$$\displaystyle \int \tan x \sin x dx+\displaystyle \int x \tan x \cos dx$$
$$\Rightarrow \ (\ln |\tan x +\sec x|-\sin x) -(x \cos x +\sin x)$$
$$x \tan x+\ln |\cos x|-(x\sin x \tan x+x \cos x-\ln |\tan x+\sec x|)$$
$$\Rightarrow \ \tan x+\ln |\cos x|-x\sin x \tan x -x\cos x +\ln (\tan x+\sec x)+C$$
$$\Rightarrow \ =\displaystyle \lim _{x+0} \to 0$$
$$=\displaystyle \lim _{x+n}\Rightarrow \ \pi$$
$$\therefore \ \pi -0\ , \ I=\pi$$

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Subjective Medium Published on 17th 09, 2020
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