If int frac{1}{1 + sin x} dx = tan(f(x)) + c, | vedclass

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Question

(C) We are given the integral $\int \frac{1}{1 + \sin x} dx$. Multiply the numerator and denominator by $(1 - \sin x)$:$\int \frac{1 - \sin x}{1 - \si

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) We are given the integral $\int \frac{1}{1 + \sin x} dx$. Multiply the numerator and denominator by $(1 - \sin x)$:$\int \frac{1 - \sin x}{1 - \sin^2 x} dx = \int \frac{1 - \sin x}{\cos^2 x} dx = \int (\sec^2 x - \sec x 	an x) dx = 	an x - \sec x + c$. Alternatively,using the half-angle identity $\sin x = \cos(\frac{\pi}{2} - x)$:$\int \frac{1}{1 + \cos(\frac{\pi}{2} - x)} dx = \int \frac{1}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} dx = \frac{1}{2} \int \sec^2(\frac{\pi}{4} - \frac{x}{2}) dx$. Integrating this,we get $\frac{1}{2} \cdot \frac{	an(\frac{\pi}{4} - \frac{x}{2})}{-\frac{1}{2}} + c = -	an(\frac{\pi}{4} - \frac{x}{2}) + c = 	an(\frac{x}{2} - \frac{\pi}{4}) + c$. Comparing this with $	an(f(x)) + c$,we have $f(x) = \frac{x}{2} - \frac{\pi}{4}$. Thus,$f'(x) = \frac{d}{dx}(\frac{x}{2} - \frac{\pi}{4}) = \frac{1}{2}$. Therefore,$f'(0) = \frac{1}{2}$.

If $\int \frac{1}{1 + \sin x} dx = \tan(f(x)) + c$,then $f'(0) =$

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