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

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Question

(A) We have the integral $I = \int e^{	an x}(\sec^2 x + \sec^3 x \sin x) dx$. Since $\sec^3 x \sin x = \sec^2 x \cdot \sec x \sin x = \sec^2 x 	an x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We have the integral $I = \int e^{	an x}(\sec^2 x + \sec^3 x \sin x) dx$. Since $\sec^3 x \sin x = \sec^2 x \cdot \sec x \sin x = \sec^2 x 	an x$,we can rewrite the integral as: $I = \int e^{	an x}(\sec^2 x + \sec^2 x 	an x) dx = \int e^{	an x} \sec^2 x (1 + 	an x) dx$. Let $u = 	an x$,then $du = \sec^2 x dx$. Substituting these into the integral,we get: $I = \int e^u (1 + u) du = \int (e^u + u e^u) du$. Using the standard integral form $\int e^u (f(u) + f'(u)) du = e^u f(u) + c$,where $f(u) = u$ and $f'(u) = 1$: $I = e^u \cdot u + c = e^{	an x} 	an x + c$.

$\int e^{\tan x}(\sec^2 x + \sec^3 x \sin x) dx =$

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