Question

Integrate $\int \left(\sqrt{\tan x}+\sqrt{\cot x}\right)\mathrm{dx}$.

Answer 1

Find the integral:

Given, $\int \left(\sqrt{\tan x}+\sqrt{\cot x}\right)\mathrm{d}x$

We know that, $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\mathrm{cosx}}{\mathrm{sinx}}$. Therefore,

$\begin{array}{c}\int \left(\sqrt{\mathrm{tanx}}+\sqrt{\mathrm{cotx}}\right)\mathrm{d}x=\int \left[\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}\right]\mathrm{d}x\\ =\int \frac{\sqrt{{\sin }^{2}x}+\sqrt{{\cos }^{2}x}}{\sqrt{\sin x·\cos x}}\mathrm{d}x\\ =\int \frac{\sin x+\cos x}{\sqrt{\sin x·\cos x}}\mathrm{d}x...\left(\mathrm{i}\right)\end{array}$

Let us assume that $\sin \left(x\right)-\cos \left(x\right)=t$

On differentiating,

$\begin{array}{cc}\Rightarrow & \frac{d}{dx}\left[\sin x-\cos x\right]=\frac{dt}{dx}\\ \Rightarrow & \left[\cos x+\sin x\right]\mathrm{d}x=\mathrm{d}t...\left(\mathrm{ii}\right)\end{array}$

From our assumption,

$\begin{array}{ccc}\Rightarrow & {\left[\sin x-\cos x\right]}^2=t^2& \\ \Rightarrow & {\sin }^{2}x+{\cos }^{2}x-2\sin x·\cos x=t^2& \\ \Rightarrow & 1-2\sin x·\cos x=t^2& \left[\because {\sin }^{2}x+{\cos }^{2}x=1\right]\\ \Rightarrow & \sin x·\cos x=\frac{1-t^2}{2}& ...\left(\mathrm{iii}\right)\end{array}$

Substituting $\left(\mathrm{iii}\right)$ and $\left(\mathrm{ii}\right)$ in $\left(\mathrm{i}\right)$,

$\begin{array}{rcl}\int \frac{\sin x+\cos x}{\sqrt{\sin x·\cos x}}\mathrm{d}x& =& \int \frac{1}{\sqrt{{\displaystyle \frac{1-t^2}{2}}}}\mathrm{d}t\\ & =& \int \frac{\sqrt{2}}{\sqrt{1-t^2}}\mathrm{d}t\\ & =& \sqrt{2}\int \frac{1}{\sqrt{1-t^2}}\mathrm{d}t\\ & =& \sqrt{2}{\sin }^{-1}\left(t\right)+\mathrm{C}\left[\because \int \frac{1}{\sqrt{1-x^2}}={\sin }^{-1}\left(x\right)\right]\end{array}$

Substituting the value of $t$ in the above equation,

$\Rightarrow \sqrt{2}{\sin }^{-1}\left(t\right)+C=\sqrt{2}{\sin }^{-1}\left(\sin x-\cos x\right)\begin{array}{l}+\end{array}C$

Therefore, $\int \left(\sqrt{\mathrm{tanx}}+\sqrt{\mathrm{cotx}}\right)\mathrm{d}x=\sqrt{2}{\sin }^{-1}\left(\sin x-\cos x\right)+\mathrm{C}$.