If int frac{dx}{1+sin x} = tan left(frac{x}{2 | vedclass

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(B) We have the integral $I = \int \frac{dx}{1+\sin x}$.Multiply the numerator and denominator by $(1-\sin x)$:$I = \int \frac{1-\sin x}{1-\sin^2 x} d

Vedclass · Accepted Answer

$\frac{\pi}{4}$

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(B) We have the integral $I = \int \frac{dx}{1+\sin x}$.Multiply the numerator and denominator by $(1-\sin x)$:$I = \int \frac{1-\sin x}{1-\sin^2 x} dx = \int \frac{1-\sin x}{\cos^2 x} dx$$I = \int (\sec^2 x - \sec x 	an x) dx = 	an x - \sec x + C$Using half-angle identities $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$ and $\cos x = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2})$:$I = \frac{\sin x - 1}{\cos x} + C = \frac{-(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))^2}{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))} + C$$I = -\frac{\cos(\frac{x}{2}) - \sin(\frac{x}{2})}{\cos(\frac{x}{2}) + \sin(\frac{x}{2})} + C = -	an\left(\frac{\pi}{4} - \frac{x}{2}ight) + C$Since $-	an(A) = 	an(-A)$,we have $I = 	an\left(\frac{x}{2} - \frac{\pi}{4}ight) + C$.Comparing this with $	an(\frac{x}{2} - 	heta) + C$,we get $	heta = \frac{\pi}{4}$.

If $\int \frac{dx}{1+\sin x} = \tan \left(\frac{x}{2}-\theta\right) + C$,then $\theta=$

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