int e^{tan x}(sec ^{2} x+sec ^{3} x sin x) d | vedclass

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Question

(B) Let $I = \int e^{	an x}(\sec ^{2} x + \sec ^{3} x \sin x) d x$. Since $\sec ^{3} x \sin x = \sec ^{2} x \cdot \sec x \sin x = \sec ^{2} x 	an x$

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$	an x e^{	an x}+c$

Vedclass · Answer

(B) Let $I = \int e^{	an x}(\sec ^{2} x + \sec ^{3} x \sin x) d x$. Since $\sec ^{3} x \sin x = \sec ^{2} x \cdot \sec x \sin x = \sec ^{2} x 	an x$,the integral becomes: $I = \int e^{	an x}(\sec ^{2} x + \sec ^{2} x 	an x) d x$. $I = \int e^{	an x}(1 + 	an x) \sec ^{2} x d x$. Let $t = 	an x$,then $dt = \sec ^{2} x d x$. Substituting these into the integral: $I = \int e^{t}(1 + t) d t$. $I = \int (e^{t} + t e^{t}) d t$. Using the integration by parts formula $\int (f(t) + f'(t)) e^{t} d t = f(t) e^{t} + c$,where $f(t) = t$ and $f'(t) = 1$: $I = t e^{t} + c$. Substituting back $t = 	an x$: $I = 	an x e^{	an x} + c$.

$\int e^{\tan x}(\sec ^{2} x+\sec ^{3} x \sin x) d x$ is equal to

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