Question

Prove that $\int\limits_0^{\dfrac{\pi }{4}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)} dx = \sqrt 2 .\dfrac{{3\pi }}{2}$.

Answer 1

Hint: In this question we have been given a trigonometric equation $\int\limits_0^{\dfrac{\pi }{4}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)} dx$ we need to prove that it is equal to $\sqrt 2 .\dfrac{{3\pi }}{2}$. For that firstly we will change the tan and the cot functions into sine and cosine, after that we will be using the identity ${\sin ^2}x + {\cos ^2}x = 1$ and ${\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}$ for simplifying the equation. After that we will put the limits and solve the equation accordingly to prove the result.

Complete step by step answer:
We have provided with a trigonometric equation $\int\limits_0^{\dfrac{\pi }{4}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)} dx$,
For solving this equation in the first step we will change the tan and cot function to sine and cosine functions, as we know $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$.
Now the equation would become $_0{\smallint ^{\dfrac{\pi }{4}}}\left( {\sqrt {\dfrac{{\sin x}}{{\cos x}}} + \sqrt {\dfrac{{\cos x}}{{\sin x}}} } \right)dx$,
Now we will be taking the LCM and the equation would be $_0{\smallint ^{\dfrac{\pi }{4}}}\left( {\dfrac{{\sin x + \cos x}}{{\sqrt {\sin x\cos x} }}} \right)dx$,

Now we will multiply and divide the equation by $\sqrt 2 $,
So, the equation would become, ${\sqrt 2 _0}{\smallint ^{\dfrac{\pi }{4}}}\left( {\dfrac{{\sin x + \cos x}}{{\sqrt {2\sin x\cos x} }}} \right)dx$,
Now we will add and subtract 1 in the denominator as it won’t make any change in the equation but then we can use that 1 to solve it further,
The equation would become, ${\sqrt 2 _0}{\smallint ^{\dfrac{\pi }{4}}}\left( {\dfrac{{\sin x + \cos x}}{{\sqrt {1 + 2\sin x\cos x - 1} }}} \right)dx$,
Now we will be using the relation, ${\sin ^2}x + {\cos ^2}x = 1$
The equation would become, ${\sqrt 2 _0}{\smallint ^{\dfrac{\pi }{4}}}\left( {\dfrac{{\sin x + \cos x}}{{\sqrt {1 - ({{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x)} }}} \right)dx$,
Now we will be using the identity ${a^2} + {b^2} - 2ab = {(a - b)^2}$,
The equation would become, ${\sqrt 2 _0}{\smallint ^{\dfrac{\pi }{4}}}\left( {\dfrac{{\sin x + \cos x}}{{\sqrt {1 - {{(\sin x - \cos x)}^2}} }}} \right)dx$,
Now we will let $\sin x - \cos x = a$
So, $da = (\cos x + \sin x)dx$
Substitute these values in the equation,
Also, the limit will also change accordingly,
As we have $\sin x - \cos x = a$,
So, when we will keep the upper limit that is $\dfrac{\pi }{4}$ in place of x we will get 0, which will be the new upper limit. Similarly, we will keep the lower limit that is 0 in place of x in the same equation we will get -1, which is our new lower limit.
So, the equation would become, ${\sqrt 2 _{ - 1}}{\smallint ^0}\left( {\dfrac{a}{{\sqrt {1 - {a^2}} }}} \right)da$,
Now we will be using the standard integral, ${\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}$,
The equation would become, $da = $$\sqrt 2 {\left[ {{{\sin }^{ - 1}}a} \right]^0}_{ - 1}$
Now applying the limits we get, $\sqrt 2 \left( {{{\sin }^{ - 1}}(0) - {{\sin }^{ - 1}}( - 1)} \right)da$,
Now simplifying the equation, $\sqrt 2 \left( {0 - \left( {\dfrac{{ - 3\pi }}{2}} \right)} \right)$,
So, the answer would be, $\sqrt 2 \dfrac{{3\pi }}{2}$,
We can also write it as $\dfrac{{3\pi }}{{\sqrt 2 }}$,
Hence, LHS=RHS.

Note:
In this question we can use the identity ${\sin ^2}x + {\cos ^2}x = 1$ in the second step only but as we need to prove $_0{\smallint ^{\dfrac{\pi }{4}}}\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)dx = \sqrt 2 .\dfrac{{3\pi }}{2}$ so we need to add and subtract 1 in the denominator for that purpose and then we can use this identity in place of 1.