int e^x left( frac{2+sin 2x}{1+cos 2x} right) | vedclass

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

Vedclass · Accepted Answer

$e^x 	an x + C$

Vedclass · Answer

(D) Let $I = \int e^x \left( \frac{2+\sin 2x}{1+\cos 2x} ight) dx$. Using the trigonometric identities $1+\cos 2x = 2\cos^2 x$ and $\sin 2x = 2\sin x \cos x$, we get: $I = \int e^x \left( \frac{2 + 2\sin x \cos x}{2\cos^2 x} ight) dx$ $I = \int e^x \left( \frac{2}{2\cos^2 x} + \frac{2\sin x \cos x}{2\cos^2 x} ight) dx$ $I = \int e^x (\sec^2 x + 	an x) dx$. We know that $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$. Here, $f(x) = 	an x$ and $f'(x) = \sec^2 x$. Therefore, $I = e^x 	an x + C$.

$\int e^x \left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx$ is equal to

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