The value of int e^{x}left[frac{1+sin x}{1+co | vedclass

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Question

(A) Let $I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} \right) dx$. Using trigonometric identities,$1+\sin x = 1+2 \sin \frac{x}{2} \cos \frac{x}{2}

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

Vedclass · Answer

(A) Let $I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} ight) dx$. Using trigonometric identities,$1+\sin x = 1+2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$. Substituting these into the integral: $I = \int e^{x} \left( \frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} ight) dx$ $I = \int e^{x} \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} ight) dx$ $I = \int e^{x} \left( \frac{1}{2} \sec^2 \frac{x}{2} + 	an \frac{x}{2} ight) dx$. Let $f(x) = 	an \frac{x}{2}$. Then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$. Since the integral is in the form $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$,we get $I = e^{x} 	an \frac{x}{2} + C$.

The value of $\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x$ is equal to

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