int {left( {frac{{2 + sin 2x}}{{1 + cos 2x}}} | vedclass

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Question

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

Vedclass · Accepted Answer

${e^x}	an x + c$

Vedclass · Answer

(D) We are given the integral $I = \int {e^x \left( \frac{2 + \sin 2x}{1 + \cos 2x} ight) dx}$.Using the trigonometric identities $1 + \cos 2x = 2\cos^2 x$ and $\sin 2x = 2\sin x \cos x$,we can simplify the integrand:$\frac{2 + \sin 2x}{1 + \cos 2x} = \frac{2 + 2\sin x \cos x}{2\cos^2 x} = \frac{2(1 + \sin x \cos x)}{2\cos^2 x} = \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} = \sec^2 x + 	an x$.Now,the integral becomes $I = \int e^x (	an x + \sec^2 x) dx$.We know the standard integral form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$.Here,$f(x) = 	an x$ and $f'(x) = \sec^2 x$.Therefore,$I = e^x 	an x + c$.

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

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