int e^x left( frac{1 + sin x}{1 + cos x} righ | vedclass

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(B) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$. Given integral is $I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx$. Using

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$e^x 	an \frac{x}{2}$

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(B) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$. Given integral is $I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} ight) dx$. Using trigonometric identities: $1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$. So,$\frac{1 + \sin x}{1 + \cos x} = \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1}{2} \sec^2 \frac{x}{2} + 	an \frac{x}{2}$. Let $f(x) = 	an \frac{x}{2}$. Then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$. Thus,the integral becomes $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C = e^x 	an \frac{x}{2} + C$. Therefore,the correct option is $B$.

$\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx = $ . . . . . . $+ C$.

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