Mathematics

The value of integral $$\displaystyle \int _{ \pi /4 }^{ 3\pi /4 }{ \frac { x }{ 1+\sin x } dx } $$ is 


ANSWER

$$\pi (\sqrt { 2 } -1)$$


SOLUTION
Let us take $$I = \displaystyle \int_{\pi/4}^{3\pi/4} \dfrac{x}{1 + \sin x} dx $$ ___(1)

we can write 

$$I = \displaystyle \int_{\pi/4}^{3 \pi/4} \dfrac{\pi - x}{1 + \sin x} dx$$ ___(2)

adding (1) and (2) we get

$$2I = \displaystyle \int_{\pi/4}^{3 \pi/4} \dfrac{\pi}{1 + \sin x} dx \Rightarrow I = \dfrac{\pi}{2} \int_{\pi/4}^{3\pi/4} \dfrac{dx}{1 + \sin x}$$

$$I = \dfrac{\pi}{2} \displaystyle \int_{\pi/4}^{3 \pi/4} \dfrac{1 - \sin x}{\cos^2 x} dx$$

$$I = \dfrac{\pi}{2} \displaystyle \int_{\pi/4}^{3\pi/4} \sec^2 x - \sec x \tan x \, dx$$

$$I = \dfrac{\pi}{2} \tan x - \sec x |_{\pi/4}^{3\pi/4}$$

$$= \dfrac{\pi}{2} (-1 + \sqrt{2} - 1 + \sqrt{2}) = \pi (\sqrt{2} - 1)$$
option (B)
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Single Correct Medium Published on 17th 09, 2020
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