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

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$ e^{x} 	an \left(\frac{x}{2}ight)+C $

Vedclass · Answer

(A) Given the integral $ I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} ight) dx $.Using trigonometric identities $ 1+\cos x = 2\cos^2 \left( \frac{x}{2} ight) $ and $ \sin x = 2\sin \left( \frac{x}{2} ight) \cos \left( \frac{x}{2} ight) $:$ I = \int e^{x} \left( \frac{1 + 2\sin \left( \frac{x}{2} ight) \cos \left( \frac{x}{2} ight)}{2\cos^2 \left( \frac{x}{2} ight)} ight) dx $$ I = \int e^{x} \left( \frac{1}{2\cos^2 \left( \frac{x}{2} ight)} + \frac{2\sin \left( \frac{x}{2} ight) \cos \left( \frac{x}{2} ight)}{2\cos^2 \left( \frac{x}{2} ight)} ight) dx $$ I = \int e^{x} \left( \frac{1}{2} \sec^2 \left( \frac{x}{2} ight) + 	an \left( \frac{x}{2} ight) ight) dx $We know that $ \int e^{x} (f(x) + f'(x)) dx = e^{x} f(x) + C $.Here,let $ f(x) = 	an \left( \frac{x}{2} ight) $. Then $ f'(x) = \sec^2 \left( \frac{x}{2} ight) \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \left( \frac{x}{2} ight) $.Thus,$ I = e^{x} 	an \left( \frac{x}{2} ight) + C $.

$ \int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x $ is

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