Question

Find $\int [\sqrt{cotx}+\sqrt{tanx}]dx$

Answer 1

$\begin{array}{c}\int \left[\sqrt{\cot x}+\sqrt{\tan x}\right]dx\Rightarrow \int \left[\sqrt{\cot x}+\frac{1}{\sqrt{\cot x}}\right]dx\\ \Rightarrow \int \left[\frac{\cot x+1}{\sqrt{\cot x}}\right]dx\Rightarrow \int \left[\sqrt{\tan x}\left(\cot x+1\right)\right]dx\\ Let\tan x=t^{2}diff.w.r.t.x,{Sec}^{2}x=2t\frac{dt}{dx}\\ \left(1+{\tan }^{2}x\right)=2t.\frac{dt}{dx}\Rightarrow 1+{\left(t^2\right)}^2=2t.\frac{dt}{dx}\\ \Rightarrow \left(1+t^4\right)dx=2tdt\Rightarrow dx=\frac{2t}{1+t^4}dt\end{array}$

Putting the value of t & dt, we get

$\begin{array}{cc}\int \left[\sqrt{t^2}\left(\cot x+1\right)\right]dx\Rightarrow \int \left[\sqrt{t^2}\left(\frac{1}{\tan x}+1\right)\right]dx\Rightarrow \int t\left[\frac{1}{t^2}+1^2\right]dx& \\ \Rightarrow \int t\left[\frac{1+t^2}{t^2}\right]dx=\int t\left(\frac{1+t^2}{t^2}\right)\times \frac{2t}{1+t^2}.dt& \\ \Rightarrow \int 2\frac{\left(1+t^2\right)}{1+t^4}dt& DivideN\&Dby{t}^2\\ \Rightarrow 2\int \frac{\frac{1}{t^2}+1}{\frac{1}{t^2}+t^2}dt& \left[Adding\&subtracting2inDenominator\right]\\ \Rightarrow 2\int \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{e}\right)+{\left(\sqrt{2}\right)}^2}dtlett-\frac{1}{t}=y& diff.weget\\ 1+\frac{1}{t^2}=\frac{dy}{dt}\Rightarrow dt=\frac{dy}{\left(1+\frac{1}{t^2}\right)}& \\ Puttingthevalueof\left(t-\frac{1}{t}\right)\&dt,weget& \\ \Rightarrow 2\int \frac{1+\frac{1}{t^2}}{y^2+{\left(\sqrt{2}\right)}^2}dt\Rightarrow 2\int \frac{1}{y^2+{\left(\sqrt{2}\right)}^2}dy& \\ \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c_1\Rightarrow 2\int \left(\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{y}{\sqrt{2}}+c_1\right)& \\ \Rightarrow \frac{2}{\sqrt{2}}{\tan }^{-1}\frac{y}{\sqrt{2}}+2c_1\Rightarrow \sqrt{2}{\tan }^{-1}\frac{\frac{1}{t}-t}{\sqrt{2}}+2c_1& \\ \Rightarrow \sqrt{2}{\tan }^{-1}x\left(\frac{\tan x-1}{\sqrt{2}\sqrt{\tan x}}\right)+c& \left[usingt=\sqrt{\tan x}\right]\\ \Rightarrow \sqrt{2}{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+c& \end{array}$