Question

Evaluate $\int\limits_0^{\dfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)dx} $.

Answer 1

Hint: As we are given two trigonometric functions inside the integral, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and now we can use the substitution method by taking $\sin x - \cos x = t$and by changing the limits accordingly and we now the formula $\int {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x} $, using which we get the final answer.

Step by step solution :
We are asked to evaluate $\int\limits_0^{\dfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)dx} $
We are given two trigonometric functions inside the integral
We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$
Using this in the given integral we get
$
   \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left( {\sqrt {\dfrac{{\sin x}}{{\cos x}}} + \sqrt {\dfrac{{\cos x}}{{\sin x}}} } \right)dx} \\
   \Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{\sin x + \cos x}}{{\sqrt {\sin x\cos x} }}} \right)dx} \\
 $
Let the above equation be (1)
Now let $\sin x - \cos x = t$
Differentiating this we get \[\left( {\cos x + \sin x} \right)dx = dt\]……..(2)
Squaring t we get
$
   \Rightarrow {t^2} = {\left( {\sin x - \cos x} \right)^2} \\
   \Rightarrow {t^2} = {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x \\
   \Rightarrow {t^2} = 1 - 2\sin x\cos x \\
   \Rightarrow 2\sin x\cos x = 1 - {t^2} \\
   \Rightarrow \sin x\cos x = \dfrac{{1 - {t^2}}}{2} \\
 $
Let's use this and equation (2) in equation (1)
Now considering the limits
When x = 0 ,
$
   \Rightarrow \sin 0 - \cos 0 = t \\
   \Rightarrow - 1 = t \\
 $
When $x = \dfrac{\pi }{2}$
$
   \Rightarrow \sin \dfrac{\pi }{2} - \cos \dfrac{\pi }{2} = t \\
   \Rightarrow 1 = t \\
 $
Therefore the limit ranges from – 1 to 1
$
   \Rightarrow \int\limits_{ - 1}^1 {\left( {\dfrac{{dt}}{{\sqrt {\dfrac{{1 - {t^2}}}{2}} }}} \right)} \\
   \Rightarrow \int\limits_{ - 1}^1 {\left( {\dfrac{{\sqrt 2 dt}}{{\sqrt {1 - {t^2}} }}} \right)} \\
 $
We know that $\int {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x} $
Using this we get
 $
   \Rightarrow \sqrt 2 \left[ {{{\sin }^{ - 1}}t} \right]_{ - 1}^1 \\
   \Rightarrow \sqrt 2 \left[ {{{\sin }^{ - 1}}\left( 1 \right) - {{\sin }^{ - 1}}\left( { - 1} \right)} \right] \\
   \Rightarrow \sqrt 2 \left[ {\dfrac{\pi }{2} - \dfrac{{3\pi }}{2}} \right] = \sqrt 2 \left( {\dfrac{{ - 2\pi }}{2}} \right) \\
   \Rightarrow - \sqrt 2 \pi \\
 $

Note :
Steps to keep in mind while solving trigonometric problems are:
1) Always start from the more complex side.
2) Express everything into sine and cosine.
3) Combine terms into a single fraction.
4) Use Pythagorean identities to transform between ${\sin ^2}x{\text{ and }}{\cos ^2}x$.
5) Know when to apply a double angle formula.
6) Know when to apply an additional formula.
7) Good old expand/ factorize/ simplify/ cancelling.
8) Integration, in mathematics, technique of finding a function g(x) the derivative of which, Dg(x), is equal to a given function f(x). This is indicated by the integral sign $\int {} $ as in $\int {f(x)} $, usually called the indefinite integral of the function.