int frac{tan ^4 sqrt{x} cdot sec ^2 sqrt{x}}{ | vedclass

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

Vedclass · Accepted Answer

$\frac{2}{5}[	an \sqrt{x}]^5+c$

Vedclass · Answer

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

$\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x=$

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