If int {frac{{{e^x}(1 + sin x)}}{{1 + cos x}} | vedclass

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(C) We are given the integral $I = \int {e^x \left( \frac{1 + \sin x}{1 + \cos x} \right)} dx$. Using the trigonometric identities $1 + \cos x = 2\cos

Vedclass · Accepted Answer

$	an \frac{x}{2}$

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(C) We are given the integral $I = \int {e^x \left( \frac{1 + \sin x}{1 + \cos x} ight)} dx$. Using the trigonometric identities $1 + \cos x = 2\cos^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$,we get: $I = \int e^x \left[ \frac{1 + 2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} ight] dx$ $I = \int e^x \left[ \frac{1}{2\cos^2(x/2)} + \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} ight] dx$ $I = \int e^x \left[ \frac{1}{2}\sec^2(x/2) + 	an(x/2) ight] dx$. Let $f(x) = 	an(x/2)$. Then $f'(x) = \frac{1}{2}\sec^2(x/2)$. Since the integral is of the form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$,we have: $I = e^x 	an(x/2) + c$. Comparing this with the given expression ${e^x}f(x) + c$,we find $f(x) = 	an(x/2)$.

If $\int {\frac{{{e^x}(1 + \sin x)}}{{1 + \cos x}}} dx = {e^x}f(x) + c$,then $f(x) = $

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