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.

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.