Mathematics

If $$\int \dfrac {1}{1 + \sin x}dx = \tan \left (\dfrac {x}{2} + a\right ) + b$$, then


ANSWER

$$a = \dfrac {\pi}{4}, b\in R$$


SOLUTION
$$\int \dfrac{1}{1+\sin x}dx= \tan \left(\dfrac{x}{2}+a\right)+b$$
differentiating both sides wrt to x, we get :
$$\dfrac{1}{1+\sin x}=\dfrac{d}{dx}\left(\tan \left(\dfrac{x}{2}+a\right)+b\right)$$

$$\therefore \dfrac{1}{1+\sin x}= \sec^{2}\left(\dfrac{x}{2}+a\right)\times \dfrac{1}{2}$$

$$\therefore \dfrac{2}{1+\sin x}=\sec^{2}\left(\dfrac{x}{2}+a\right)$$

$$\therefore 1+\sin x=2 \cos^{2}\left(\dfrac{x}{2}+a\right)$$

$$\therefore \sin x= \cos (x+2a)$$

$$\therefore \cos \left(\dfrac{\pi }{2}-x\right)=\cos (x+2a)$$

$$\therefore \dfrac{\pi }{2}-2a=2x$$

$$\therefore \dfrac{\pi }{2}-2a=0$$

$$\therefore a^{2}=\dfrac{\pi }{2\times 2}=\dfrac{\pi }{4}$$

$$\therefore a=\dfrac{\pi }{4},b=R$$
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Single Correct Medium Published on 17th 09, 2020
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