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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } I=\int \frac{	an ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x \ & 	ext { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \ & 	herefore I=2 \int 	an ^4 \operatorname{tsec}^2 t d t \end{aligned}$ Put $	an \mathrm{t}=\mathrm{u} \quad \Rightarrow \sec ^2 \mathrm{tdt}=\mathrm{du}$ $\begin{aligned} & 	herefore \mathrm{I}=2 \int \mathrm{u}^4 \mathrm{du} \ & =\frac{2 \mathrm{u}^5}{5}=\frac{2(	an \mathrm{t})^5}{5}=\frac{2}{5} 	an ^5 \sqrt{\mathrm{x}}+\mathrm{c} \end{aligned}$

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

Solution