Question

$\int \left[\frac{\sqrt{\tan x}}{\left(\sin x\cos x\right)}\right]dx=\_\_\_\_\_\_+c$; $x\ne \frac{k\mathrm{\pi }}{2}$ and $\tan x>0$

• A. $\frac{1}{2}\sqrt{\tan x}$
• B. $\sqrt{\left(2\tan x\right)}$
• C. $2\sqrt{\left(\tan x\right)}$
• D. $\sqrt{\tan x}$

Answer 1

The correct option is C

$2\sqrt{\left(\tan x\right)}$

Explanation for the correct option:

Determine the value of $\int \left[\frac{\sqrt{\tan x}}{\left(\sin x\cos x\right)}\right]dx$

Consider the given Equation as

$I=\int \left[\frac{\sqrt{\tan x}}{\left(\sin x\cos x\right)}\right]dx$

Multiply and divided by ${\cos }^{2}x$ in the denominator of the above Equation

$$\begin{gathered} I=\int \left[\frac{\sqrt{\tan x}}{\left({\displaystyle \frac{\sin x\cos x}{{\cos }^{2}x}}\right){\cos }^{2}x}\right]dx \\ I=\int \left[\frac{\sqrt{\tan x}}{\left({\displaystyle \frac{\sin x}{\cos x}}\right){\cos }^{2}x}\right]dx \\ I=\int \left[\frac{\sqrt{\tan x}}{\left(\tan x\right)\left({\cos }^{2}x\right)}\right]dx\left(\text{where},\frac{\sin x}{\cos x}=\tan x\right) \\ I=\int \left[\frac{\sqrt{\tan x}}{\left(\tan x\right)}se{c}^{2}x\right]dx\left(\text{where},\frac{1}{\left({\cos }^{2}x\right)}=se{c}^{2}x\right) \\ I=\int \left[\frac{{\left(\tan x\right)}^{{\displaystyle \frac{1}{2}}}}{\left(\tan x\right)}se{c}^{2}x\right]dx \\ I=\int \left[\frac{1}{{\left(\tan x\right)}^{\frac{1}{2}}}se{c}^{2}x\right]dx \end{gathered}$$

Let us assume that

$$\begin{gathered} \tan x=t \\ \text{Differentiate with respect to}x \\ se{c}^{2}xdx=dt \end{gathered}$$

Substitute the above values in the integral

$$\begin{gathered} I=\int t^{-\frac{1}{2}}dt \\ I=\frac{t^{-\frac{1}{2}+1}}{\left(-{\displaystyle \frac{1}{2}}+1\right)}+c \\ I=\frac{t^{\frac{1}{2}}}{{\displaystyle \frac{1}{2}}}+c \\ I=2\left(t^{\frac{1}{2}}\right)+c \\ I=2{\left(\tan x\right)}^{\frac{1}{2}}+c\left(\text{Where},t=\tan x\right) \\ I=2\sqrt{\left(\tan x\right)}+c \end{gathered}$$

Hence, the correct answer is Option C.