int e^{-2 x}left(tan 2 x-2 sec ^2 2 x tan 2 x | vedclass

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(C) Let $I = \int e^{-2 x}(	an 2 x - 2 \sec^2 2 x 	an 2 x) dx$. Consider the function $f(x) = 	an 2 x$. Then $f'(x) = 2 \sec^2 2 x$. The integral i

Vedclass · Accepted Answer

$-\frac{e^{-2 x}}{2}\left[	an 2 x-\sec ^2 2 xight]+c$

Vedclass · Answer

(C) Let $I = \int e^{-2 x}(	an 2 x - 2 \sec^2 2 x 	an 2 x) dx$. Consider the function $f(x) = 	an 2 x$. Then $f'(x) = 2 \sec^2 2 x$. The integral is of the form $\int e^{ax} (f(x) + \frac{1}{a} f'(x)) dx$ is not directly applicable here due to the term $	an 2 x$. Let $2x = t$, then $2 dx = dt$, so $dx = \frac{1}{2} dt$. $I = \frac{1}{2} \int e^{-t} (	an t - 2 \sec^2 t 	an t) dt$. Let $f(t) = 	an t - \sec^2 t$. Then $f'(t) = \sec^2 t - 2 \sec^2 t 	an t$. Thus, $I = \frac{1}{2} \int e^{-t} (f'(t) - f(t)) dt$. Using the formula $\int e^{kt} (f'(t) + k f(t)) dt = e^{kt} f(t) + C$, with $k = -1$: $I = \frac{1}{2} (-e^{-t} f(t)) + C = -\frac{1}{2} e^{-t} (	an t - \sec^2 t) + C$. Substituting $t = 2x$: $I = -\frac{1}{2} e^{-2 x} (	an 2 x - \sec^2 2 x) + C = \frac{1}{2} e^{-2 x} (\sec^2 2 x - 	an 2 x) + C$.

$\int e^{-2 x}\left(\tan 2 x-2 \sec ^2 2 x \tan 2 x\right) d x=$

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