Question

Integrate the function $\int {\dfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx = \ldots .. + c$ ; $x \ne \dfrac{{k\pi }}{2}$ and $\tan x > 0$
$\left( 1 \right)$ $\dfrac{1}{{2\sqrt {\tan x} }}$
$\left( 2 \right)$ $\sqrt {2\tan x}$
$\left( 3 \right)$ $2\sqrt {\tan x}$
$\left( 4 \right)$ $\sqrt {\tan x}$

Answer 1

We are going to solve this question using integration by substitution method and using the various formulas of trigonometric functions . First we change the terms of the integration by using the formula of $\tan x$ in terms of $\sin x$ and $\cos x$ and then we will simplify the integrating the terms and then we will substitute the value of the denominator as another element and then using the various formulas of integration of trigonometric terms we will find the value of the given integral function .

Complete answer: Given : $\int {\dfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx$
Let $I = \int {\dfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx$
Now , We have to integrate $I$ with respect to ‘ $x$ ’
As , we know that
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
Substituting value of $\tan x$ in $I$ , we get the trigonometric integral as :
$I = \int {\dfrac{{\sqrt {\dfrac{{\sin x}}{{\cos x}}} }}{{\sin x\cos x}}} dx$
On simplifying the terms , we get the expression as :
$I = \int {\dfrac{1}{{\sqrt {\sin x} {{\cos }^{\dfrac{3}{2}}}x}}} dx$
Multiplying both the numerator and denominator by $\sqrt {\cos x}$ , we can write the expression as :
$I = \int {\dfrac{1}{{\sqrt {\dfrac{{\sin x}}{{\cos x}}} {{\cos }^2}x}}} dx$
We also know that the cosine function in terms of secant is given as :
$\cos x = \dfrac{1}{{\sec x}}$
Using the relation of $\cos x$ in terms of $\sec x$ and the relation of $\tan x$ in terms of $\sin x$ and $\cos x$ , we can write the expression as :
$I = \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {\tan x} }}} dx$
Now we will substitute the value of the integral as :
$\tan x = t$
As we know that the derivative of the trigonometric function is given as :
$\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$
Using the derivative of $\tan x$ .
On differentiate $t$ with respect to $x$ , we get
${\sec ^2}xdx = dt$
Substituting the value , we get the expression for integral as :
$I = \int {\dfrac{1}{{\sqrt t }}dt}$
Now , we also know that the formula of integration of ${x^n}$ is given as :
${x^n} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$
Using the formula of integration , we get the value of the integral as :
$I = 2\sqrt t + c$
Where $c$ is the integral constant .
Substituting the value of $t$ back into the integral , we get the value of the expression as :
$I = 2\sqrt {\tan x} + c$
As , we our given that the value of the integral is given as $\int {\dfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx = \ldots .. + c$ .
Hence , the value of the integral $\int {\dfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx$ is $2\sqrt {\tan x}$ .
Thus the correct option is $\left( 3 \right)$ .

Note:
As the question was of indefinite integral that’s why we added an integral constant ‘ $c$ ’ to the integration . The final answer of the integral does not consist of the integral constant as we were given the integral constant already in the question .If the question would have been of definite integral then we would not have added the integral constant to the final answer .