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

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(D) 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 - \si

Vedclass · Accepted Answer

$a = -\frac{\pi}{4}, b = 	ext{arbitrary constant}$

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(D) 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$$I = \int (\sec^2 x - \sec x 	an x) dx$$I = 	an x - \sec x + C$Using half-angle formulas $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$:$I = \int \frac{dx}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2} = \int \sec^2 \left( \frac{x}{2} + \frac{\pi}{4} ight) \cdot \frac{1}{2} dx$ is not the direct path,let's use:$I = \int \frac{dx}{1 + \cos(\frac{\pi}{2} - x)} = \int \frac{dx}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \frac{1}{2} \int \sec^2(\frac{\pi}{4} - \frac{x}{2}) dx$$I = \frac{1}{2} \cdot \frac{	an(\frac{\pi}{4} - \frac{x}{2})}{-1/2} + C = -	an(\frac{\pi}{4} - \frac{x}{2}) + C$$I = 	an(\frac{x}{2} - \frac{\pi}{4}) + C$Comparing this with $	an(\frac{x}{2} + a) + b$,we get $a = -\frac{\pi}{4}$ and $b$ is an arbitrary constant.

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

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