Mathematics

The value of integral  $$\int _ { \pi  / 4 } ^ { 3 \pi  / 4 } \dfrac { x } { 1 + \sin x } d x$$  is :


ANSWER

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


SOLUTION
$$\\I=\int_{(\tfrac{\pi}{4})}^{(\tfrac{3\pi}{4})}(\dfrac{x}{1+sinx})dx$$



$$=\int_{(\tfrac{\pi}{4})}^{(\tfrac{3\pi}{4})}(\dfrac{\pi-x}{1+sinx})dx$$



$$I=\pi\int_{(\tfrac{\pi}{4})}^{(\tfrac{3\pi}{4})}(\dfrac{1}{1+sinx})dx-I\\or\>2I=\pi\int_{(\dfrac{\pi}{4})}^{(\dfrac{3\pi}{4})}(\dfrac{1-sinx}{1^2-sin^2x})dx$$



$$I=(\tfrac{\pi}{2})\int_{(\tfrac{\pi}{4})}^{(\tfrac{3\pi}{4})}\left((\dfrac{1}{cos^2x})-(\dfrac{sinx}{cos^2x})\right)dx$$



$$=(\dfrac{\pi}{2})\int_{(\tfrac{\pi}{4})}^{(\tfrac{3\pi}{4})}(sec^2x-tanx\>secx)dx$$



$$=(\dfrac{\pi}{2})\left[tanx-secx\right]_{\pi/4}^{3\pi/4}$$



$$=(\dfrac{\pi}{2})\left[\left(tan(\dfrac{3\pi}{4})-sec(\dfrac{3\pi}{4})\right)-\left(tan(\dfrac{\pi}{4})-sec(\dfrac{\pi}{4})\right)\right]$$



$$=(\dfrac{\pi}{2})[(-1+\sqrt{2})-(1-\sqrt{2})]\\=(\dfrac{\pi}{2})(2\sqrt{2})-2$$

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

 

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Single Correct Medium Published on 17th 09, 2020
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