Question

Find the value of $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}}dx$.
A. $\pi $
B. $\dfrac{\pi }{2}$
C. $\dfrac{\pi }{4}$
D. $\dfrac{\pi }{3}$

Answer 1

Hint: In this question, we have $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}}dx$. So, we will consider this as equation 1. Now we will use the identity $\int\limits_{0}^{a}{f(x)dx=\int\limits_{0}^{a}{f(x-a)dx}}$. After that, we will get equation 2. Now we will add equations 1 and 2 to get the simplified form of the given expression. At last, we will obtain the final result by substituting the values of the limits.


Formula Used: In this question, we will use $\int\limits_{0}^{a}{f(x)dx=\int\limits_{0}^{a}{f(x-a)dx}}$.

Complete- step by -step solution: Let us consider $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}}dx$ ….... (1)
First, we will use the identity $\int\limits_{0}^{a}{f(x)dx=\int\limits_{0}^{a}{f(x-a)dx}}$
On substituting the identity in the given integral, we get
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cot \left( \dfrac{\pi }{2}-x \right)}}{\sqrt{\cot \left( \dfrac{\pi }{2}-x \right)}+\sqrt{\tan \left( \dfrac{\pi }{2}-x \right)}}}dx$
By using the Cofunction identities $\cot \left( \dfrac{\pi }{2}-x \right)=\tan x$and $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$, we get
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}}dx$ ….... (2)
Now, adding equations (1) and (2),
$\begin{align}
  & I+I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}}dx+\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}}dx \\
 & \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cot x}+\sqrt{\tan x}}{\sqrt{\cot x}+\sqrt{\tan x}}}dx \\
\end{align}$
By canceling the same terms, we get
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{(1)dx}$
We will further integrate 1 with respect to $x$.
$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$
Substitute the values of the limit, and we get
$\begin{align}
  & \Rightarrow 2I=\left[ \dfrac{\pi }{2}-0 \right] \\
 & \Rightarrow I=\dfrac{\pi }{2\times 2} \\
 & \therefore I=\dfrac{\pi }{4} \\
\end{align}$
Thus, the obtained value will be the required value of the given integration.

Option ‘C’ is correct

Note: As the functions change, it is not essential to integrate taking limits directly $0$ to $\dfrac{\pi }{2}$. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier. And it should be remembered that, on applying limits, the upper limit is applied first and the lower limit is applied.