Integral Exercise

Exercise:

Integrate, $\int{\frac{{{\tan }^{3}}x}{\sqrt{\sec x}}dx}$

Solution: we have ${{\tan }^{3}}x={{\tan }^{2}}x\tan x=\left( {{\sec }^{2}}x-1 \right)\tan x$

So $\int{\frac{{{\tan }^{3}}x}{\sqrt{\sec x}}dx}=\int{\frac{\left( {{\sec }^{2}}x-1 \right)\tan x}{\sqrt{\sec x}}dx}=\int{\left( {{\sec }^{2}}x-1 \right)\tan x{{\sec }^{-1/2}}dx}$

$=\int{{{\sec }^{3/2}}x\tan x\,dx\,\,-\int{\tan x{{\sec }^{-1/2}}}}x\,dx$

$=\int{{{\sec }^{1/2}}x\sec x\tan x\,dx-\int{\tan x\sec x{{\sec }^{-3/2}}\,dx}}$

Let $u=\sec x\Leftrightarrow du=\sec x\tan xdx$

$=\int{{{\sec }^{1/2}}x\,d\left( \sec x \right)}-\int{{{\sec }^{-3/2}}x\,d\left( \sec x \right)}$

$=\frac{2}{3}{{\sec }^{3/2}}x+2{{\sec }^{-1/2}}x+c=\frac{2}{3}\sqrt{{{\sec }^{3}}x}+\frac{2}{\sqrt{\sec x}}\,+c$