Question

Integrate the following functions.
$\int \frac{\sqrt{tanx}}{sinx.cosx}dx.$

Answer 1

$\int \frac{\sqrt{tanx}}{sinx.cosx}dx=\int \frac{\sqrt{tanx}}{sinxcosx\times \frac{cosx}{cosx}}dx$
[Multiply by $1=\frac{cosx}{cosx}$ in denominator]
$=\int \frac{\sqrt{tanx}}{tanx.co{s}^{2}x}dx=\int \frac{\sqrt{tanx}}{tanx}.se{c}^{2}xdx$
Let $tanx=t\Rightarrow se{c}^{2}x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{se{c}^{2}x}$
$\therefore \int \frac{\sqrt{tanx}}{tanx}.se{c}^{2}xdx=\int \frac{\sqrt{t}}{t}.se{c}^{2}x\frac{dt}{se{c}^{2}x}=\int \frac{1}{\sqrt{t}}dt=\int t^{-\frac{1}{2}}dt=\frac{t^{-\frac{1}{2}+1}}{(-\frac{1}{2}+1)}+C=2\sqrt{t}+C=2\sqrt{tanx}+C$