Question

Solve $\int \frac{{\sec }^{4}x}{\sqrt{\left(\tan x\right)}}dx$

• A. $\frac{2}{5}\sqrt{\tan x}(5+{\tan }^{2}x).$
• B. $\frac{2}{5}\sqrt{\tan x}(5-{\tan }^{2}x).$
• C. $\frac{2}{7}\sqrt{\tan x}(5+{\tan }^{2}x).$
• D. $\frac{2}{7}\sqrt{\tan x}(5+{\tan }^{3}x).$

Answer 1

The correct option is A $\frac{2}{5}\sqrt{\tan x}(5+{\tan }^{2}x).$

$I=\int \frac{{\sec }^{2}x{\sec }^{2}xdx}{\sqrt{\left(\tan x\right)}}.$ Put $\tan x=t.\therefore {\sec }^{2}xdx=dt.$ $\therefore I=\int \frac{1+t^2}{\sqrt{\left(t\right)}}dt=\int (\frac{1}{\sqrt{\left(t\right)}}+t^{3/2})dt.$ $=2\sqrt{t}+\frac{2}{5}{t}^{5/2}=\frac{2}{5}\sqrt{t}(5+t^2)=\frac{2}{5}\sqrt{\tan x}(5+{\tan }^{2}x).$