int frac{1}{sqrt{x}} tan^4(sqrt{x}) sec^2(sqr | vedclass

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Question

(C) Let $I = \int \frac{1}{\sqrt{x}} 	an^4(\sqrt{x}) \sec^2(\sqrt{x}) \, dx$. Substitute $t = 	an(\sqrt{x})$. Then,$dt = \sec^2(\sqrt{x}) \cdot \fra

Vedclass · Accepted Answer

$\frac{2}{5} 	an^5(\sqrt{x}) + c$

Vedclass · Answer

(C) Let $I = \int \frac{1}{\sqrt{x}} 	an^4(\sqrt{x}) \sec^2(\sqrt{x}) \, dx$. Substitute $t = 	an(\sqrt{x})$. Then,$dt = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \, dx$,which implies $\frac{\sec^2(\sqrt{x})}{\sqrt{x}} \, dx = 2 \, dt$. Substituting these into the integral,we get: $I = \int t^4 \cdot 2 \, dt = 2 \int t^4 \, dt$. Integrating $t^4$ with respect to $t$,we get $2 \cdot \frac{t^5}{5} + c$. Substituting back $t = 	an(\sqrt{x})$,we obtain $I = \frac{2}{5} 	an^5(\sqrt{x}) + c$.

$\int \frac{1}{\sqrt{x}} \tan^4(\sqrt{x}) \sec^2(\sqrt{x}) \, dx = $

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