int {{e^{2x}}frac{{1 + sin 2x}}{{1 + cos 2x}} | vedclass

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Question

(A) We have the integral $I = \int {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}} \,dx$. Using the trigonometric identities $1 + \cos 2x = 2\cos^2 x$ a

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$\frac{1}{2}{e^{2x}}	an x + c$

Vedclass · Answer

(A) We have the integral $I = \int {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}} \,dx$. Using the trigonometric identities $1 + \cos 2x = 2\cos^2 x$ and $\sin 2x = 2\sin x \cos x$,we get: $I = \int {{e^{2x}}\frac{{1 + 2\sin x \cos x}}{{2\cos^2 x}}} \,dx$ $I = \int {{e^{2x}}\left( \frac{1}{2\cos^2 x} + \frac{2\sin x \cos x}{2\cos^2 x} ight)} \,dx$ $I = \int {{e^{2x}}\left( \frac{1}{2}\sec^2 x + 	an x ight)} \,dx$ Let $f(x) = 	an x$,then $f'(x) = \sec^2 x$. The integral is of the form $\int e^{ax} [f(x) + \frac{1}{a}f'(x)] \,dx = \frac{1}{a} e^{ax} f(x) + c$. Here $a = 2$,so $I = \frac{1}{2} e^{2x} 	an x + c$.

$\int {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}} \,dx = $

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