Question

Evaluate $ \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)dx} $

Answer 1

Hint: To solve the above expression, we will use the concept of definite integral. Every definite integral has a solution with a unique value. Definite integral is expressed as $ \int\limits_a^b {f\left( x \right)dx} $ .
Where, $ a $ is the lower limit and $ b $ is the upper limit.
The definite integral is given as:
 $ \begin{array}{c}
\int\limits_a^b {f\left( x \right)dx} = \left[ {F\left( x \right)} \right]_a^b\\
 = F\left( b \right) - F\left( a \right)
\end{array} $
We will also use the following property of the definite integral:
 $ \int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx $
We will use the above relation to evaluate the integral in a simpler form. Also property of logarithm is used to get the results.

Complete step-by-step answer:
Given: The given integral is $ \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)} dx $ .
 We will assume $ I $ as the integral of $ \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)} dx $ .
 $ I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)} dx $
We will use the property of definite integral which is given as,
 $ \int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx $
In the given expression we have $ \dfrac{\pi }{2} $ for $ a $ and $ \tan x $ for $ f\left( x \right) $ . So we will substitute these values in the above property.
\[\begin{array}{l}
I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)} dx\\
I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left\{ {\tan \left( {\dfrac{\pi }{2} - x} \right)} \right\}} dx
\end{array}\]
We know that $ \tan \left( {\dfrac{\pi }{2} - x} \right) $ equals to $ \cot x $ . We substitute $ \cot x $ for $ \tan \left( {\dfrac{\pi }{2} - x} \right) $ in the above expression.
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\cot x} \right)dx} \]
We will rewrite the above expression in the $ \tan x $ form since $ \cot x $ is the inverse of $ \tan x $ .
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\log {{\left( {\tan x} \right)}^{ - 1}}dx} \]
Since, from the property of log we know that $ \log {m^n} = n\log m $ , We will apply this property in the above expression.
\[I = - \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)dx} \]
But according to our assumption, we have assumed $ I $ for \[\int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)dx} \].
On substituting $ I $ for \[\int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)dx} \]in the above expression we will get,
 $ \begin{array}{c}
I = - I\\
2I = 0\\
I = 0
\end{array} $
Hence, the value of integral \[\int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)dx} \] is 0.

Note: In order to evaluate the integral, always use the property of definite integrals. There are a certain number of properties of definite integrals which can be used to simplify the problem. This problem can also be solved by adding \[\int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\cot x} \right)dx} \] for $ I $ and \[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x} \right)dx} \] for $ I $ , which will give \[2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\tan x + \cot x} \right)dx} \] , Under the given limits the value will be 0.