int_{0}^{frac{pi}{4}} e^{tan^2 theta} sin^2 t | vedclass

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Question

(A) Let $I = \int_{0}^{\frac{\pi}{4}} e^{	an^2 	heta} \sin^2 	heta 	an 	heta d	heta$. Since $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	h

Vedclass · Accepted Answer

$\frac{1}{2} \left( \frac{e}{2} - 1 \right)$

Vedclass · Answer

(A) Let $I = \int_{0}^{\frac{\pi}{4}} e^{	an^2 	heta} \sin^2 	heta 	an 	heta d	heta$. Since $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta}$,we have $I = \int_{0}^{\frac{\pi}{4}} e^{	an^2 	heta} \frac{	an^3 	heta}{1 + 	an^2 	heta} d	heta$. Let $	an^2 	heta = u$,then $2 	an 	heta \sec^2 	heta d	heta = du$. Since $\sec^2 	heta = 1 + 	an^2 	heta = 1 + u$,we have $d	heta = \frac{du}{2 	an 	heta (1 + u)} = \frac{du}{2 \sqrt{u} (1 + u)}$. Substituting these into the integral: $I = \int_{0}^{1} e^u \frac{u^{3/2}}{1 + u} \frac{du}{2 \sqrt{u} (1 + u)} = \frac{1}{2} \int_{0}^{1} \frac{u e^u}{(1 + u)^2} du$. Using integration by parts,let $f(u) = u e^u$ and $g'(u) = (1 + u)^{-2}$,so $f'(u) = (u+1)e^u$ and $g(u) = -(1+u)^{-1}$. $I = \frac{1}{2} \left[ -\frac{u e^u}{1 + u} ight]_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{(u+1)e^u}{1+u} du$. $I = \frac{1}{2} \left( -\frac{e}{2} + 0 ight) + \frac{1}{2} \int_{0}^{1} e^u du$. $I = -\frac{e}{4} + \frac{1}{2} [e^u]_{0}^{1} = -\frac{e}{4} + \frac{1}{2} (e - 1) = \frac{e}{4} - \frac{1}{2} = \frac{1}{2} \left( \frac{e}{2} - 1 ight)$.

$\int_{0}^{\frac{\pi}{4}} e^{\tan^2 \theta} \sin^2 \theta \tan \theta d\theta =$

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