Question

Solve $\int (\sqrt{\tan x}+\sqrt{\cot x})dx$.

• A. $I=\sqrt{2}{\tan }^{-1}\left(\frac{1+\tan x}{\sqrt{2\tan x}}\right)+C_1$
• B. $I=\sqrt{2}{\tan }^{-1}\left(\frac{1-\tan x}{\sqrt{2\tan x}}\right)+C_1$
• C. $I={\tan }^{-1}\left(\frac{1-\tan x}{\sqrt{2\tan x}}\right)+C_1$
• D. None of these

Answer 1

Answer: (B)

$I=\sqrt{2}{\tan }^{-1}\left(\frac{1-\tan x}{\sqrt{2\tan x}}\right)+C_1$

We have,

$\int [\sqrt{\tan x}+\sqrt{\cot x}]dx$

$=\int [\frac{1}{\sqrt{\mathrm{cotx}}}+\sqrt{\cot x}]dx$

$=\int \frac{\cot x+1}{\sqrt{\cot x}}dx$

$=\int \sqrt{\tan x}(\cot x+1)dx$

Let $t^2=\tan x$

On differentiating this equation with respect to $x$, we get

${\sec }^{2}xdx=2tdt$

$1+{\tan }^{2}x=2t\frac{dt}{dx}$

$2t\frac{dt}{dx}=1+{\left(t^2\right)}^2$

$2t\frac{dt}{dx}=1+t^4$

$dx=\frac{2t}{1+t^4}dt$

Put the value of $t\&dt$ and we get,

$\int \left[\sqrt{t^2}(\cot x+1)\right]dx$

$=\int \left[\sqrt{t^2}(\frac{1}{\tan x}+1)\right]dx$

$=\int t[\frac{1}{t^2}+1]dx$

$=\int t\left[\frac{1+t^2}{t^2}\right]dx$

$=\int t\left[\frac{1+t^2}{t^2}\right]\times \frac{2t}{1+t^4}dt$

$=2\int \left[\frac{1+t^2}{1+t^4}\right]dt$

On dividing numerator and denominator by $t^2$, we get,

$=2\int \frac{\frac{1+t^2}{t^2}}{\frac{1+t^4}{t^2}}.dt$

$=2\int \frac{\frac{1}{t^2}+1}{\frac{1}{t^2}+t^2}dt$

$=2\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-2+2}dt$

$=2\int \frac{1+\frac{1}{t^2}}{{(t+\frac{1}{t})}^2+2}dt$

$=2\int \frac{1+\frac{1}{t^2}}{{(t+\frac{1}{t})}^2+{\left(\sqrt{2}\right)}^2}dt$

Let $t-\frac{1}{t}=y$

On differentiating both side with respect to $x$ and we get,

$1+\frac{1}{t^2}=\frac{dy}{dt}$

$dt=\frac{dy}{(1+\frac{1}{t^2})}$

On putting the value of $dt$ and $(\frac{1}{t}-t)$ and we get,

$=2\int \frac{1+\frac{1}{t^2}}{y^2+{\left(\sqrt{2}\right)}^2}\times \frac{dy}{(1+\frac{1}{t^2})}$

$=2\int \frac{1}{y^2+{\left(\sqrt{2}\right)}^2}dy$

We know that,

$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$

Then,

$=2(\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{y}{\sqrt{2}}+C)$

$=\frac{2}{\sqrt{2}}{\tan }^{-1}\frac{y}{\sqrt{2}}+2C$

$=\sqrt{2}{\tan }^{-1}\frac{y}{\sqrt{2}}+2C$

$=\sqrt{2}{\tan }^{-1}\frac{\frac{1}{t}-t}{\sqrt{2}}+2C$

$=\sqrt{2}{\tan }^{-1}\frac{1-t^2}{\sqrt{2}t}+2C$

$=\sqrt{2}{\tan }^{-1}\left(\frac{1-\tan x}{\sqrt{2}\sqrt{\tan x}}\right)+C_1\therefore C_1=2C$

$I=\sqrt{2}{\tan }^{-1}\left(\frac{1-\tan x}{\sqrt{2\tan x}}\right)+C_1$

Hence, this is the answer.