Question

Evaluate the following: $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{dx}{1+\sqrt{\cot x}}}$.

Answer 1

In order to solve this question, let suppose that $I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{dx}{1+\sqrt{\cot x}}}$and then substitute $\cot x=\dfrac{\cos x}{\sin x}$ in the expression. After expressing $I$ in terms of $\cos x$ and $\sin x$, by the property of definite integral $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(b+a-x)dx}}$, substitute $\left( \dfrac{\pi }{3}+\dfrac{\pi }{6}-x \right)$ in place of $x$ in $I$ and simplify the expression. And then add the two expressions of $I$, and get the final expression. Finally integrate the resultant expression and evaluate the answer.

Complete step by step answer:
It is given to evaluate the definite integral $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{dx}{1+\sqrt{\cot x}}}$.
Now, let us suppose that $I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{dx}{1+\sqrt{\cot x}}}$
Now as we know that $\cot x=\dfrac{\cos x}{\sin x}$, substituting the value of $\cot x$ in terms of $\cos x$ and $\sin x$ in the above expression, we get
$\Rightarrow I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{dx}{1+\sqrt{\dfrac{\cos x}{\sin x}}}}$
After further simplification of the above expression, we get
$\Rightarrow I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx......................(1)$
Now we know the property of the definite integral that $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(b+a-x)dx}}$, so extending that property in the above integral, we are substituting $\left( \dfrac{\pi }{3}+\dfrac{\pi }{6}-x \right)$ i.e. $\left( \dfrac{\pi }{2}-x \right)$ in place of $x$
$\Rightarrow I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}}{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}+\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}}}dx$
Simplifying the above integral further, we get
$\Rightarrow I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx......................(2)$
Adding equation (1) and equation (2), we get
$\Rightarrow 2I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx$
As the numerator and denominator are equal so it will be equal to 1, so
$\Rightarrow 2I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{{}}dx$
We know that integral of 1 is $x$ i.e. $\int{dx=x}$, so applying this in the above expression, we get
$\Rightarrow 2I=\left[ x \right]_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}=\left[ \dfrac{\pi }{3}-\dfrac{\pi }{6} \right]=\dfrac{\pi }{6}$
Dividing both sides by 2, we get
$\Rightarrow I=\dfrac{\pi }{12}$

Here, $I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{dx}{1+\sqrt{\cot x}}}=\dfrac{\pi }{12}$
Hence, from the above relation we can say that $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{\dfrac{dx}{1+\sqrt{\cot x}}}=\dfrac{\pi }{12}$.
So, the answer of the above integral given in question is $\dfrac{\pi }{12}$.

Note:
This is an interesting question of definite integral where the tricky part is to apply which property of definite integral. In this question, there are two things to note, the first one is the sum of upper limit and lower limit which in our case is equal to $\dfrac{\pi }{2}$ which is a complementary angle. Another thing to notice in the question is when we substituted $\cot x=\dfrac{\cos x}{\sin x}$, we got a complementary pair that is $\sin x$ and $\cos x$ in the denominator. So, students whenever you get these two things in a definite integral question apply $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(b+a-x)dx}}$ in the integral. It will simplify your integral and you will be able to evaluate the answer easily.