Mathematics

$$\displaystyle \underset{0}{\overset{\pi}{\int}} \dfrac{x \, dx}{1 + \sin \, x} = $$


ANSWER

$$\pi$$


SOLUTION
$$\int \dfrac{x}{1+\sin x}dx$$

$$\int \dfrac{x \sec x}{\tan x+\sec x}dx$$

$$=-\dfrac{x}{\tan x+\sec x}-\int \dfrac{1}{-\tan x-\sec x}dx$$

$$=-\dfrac{x}{\tan x+\sec x}+\int \dfrac{1}{\tan x+\sec x}dx$$

$$=-\dfrac{x}{\tan x+\sec x}+\int \dfrac{\cos x}{1+\sin x}dx$$
--------------------------------------------------------
substitute $$u=\sin x\rightarrow du=\cos x dx$$

$$=\int \dfrac{1}{u+1}du$$

$$=\ln (u+1)$$

$$=\ln (\sin x+1)$$

-----------------------------------------------------------------
$$=-\dfrac{x}{\tan x+\sec x}+\ln (\sin x+1)+C$$

$$\int_{0}^{\pi} \dfrac{x}{1+\sin x}dx$$

$$=\left [  -\dfrac{x}{\tan x+\sec x}+\ln (\sin x+1)\right ]_0^{\pi}$$

$$=\pi -0 $$

$$=\pi$$
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Single Correct Medium Published on 17th 09, 2020
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