Question

Which of the below is the value of $\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}$ equal to
A. $\dfrac{\pi }{8}{{\log }_{e}}2$
B. $\dfrac{\pi }{4}{{\log }_{e}}e$
C. $\dfrac{\pi }{4}{{\log }_{e}}2$
D. $\dfrac{\pi }{8}{{\log }_{e}}\left( \dfrac{1}{2} \right)$

Answer 1

Hint: To find the value of given integral we will use property of definite integral. Firstly we will use the property of definite integral to simplify our value inside the integral sign. Then we will use the tangent function formula to expand the term inside. Finally we will use logarithm property and solve the obtained value and get the desired answer.

Complete step by step answer:
We have to find the value of:
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}$……$\left( 1 \right)$
Using the below definite integral property:
$\int\limits_{0}^{a}{f\left( x \right)dx=\int\limits_{0}^{a}{f\left( a-x \right)dx}}$
Using above property in (1) we get,
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan \left( \dfrac{\pi }{4}-x \right) \right)dx}$……$\left( 2 \right)$
We know tangent formula given as:
$\tan \left( \dfrac{\pi }{4}-x \right)=\dfrac{1-\tan x}{1+\tan x}$
Using it in equation (2) we get,
$\begin{align}
  & I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\dfrac{1-\tan x}{1+\tan x} \right)dx} \\
 & I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( \dfrac{1+\tan x+1-\tan x}{1+\tan x} \right)dx} \\
 & I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( \dfrac{1}{1+\tan x} \right)dx} \\
\end{align}$
Now we will use logarithm property above which states that:
$\log \left( \dfrac{a}{b} \right)=\log a-\log b$
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx-\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}}$
We can replace second term from equation (1) and get,
$\begin{align}
  & I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx-I} \\
 & I+I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx} \\
 & 2I=\log 2\int\limits_{0}^{\dfrac{\pi }{4}}{1dx} \\
 & 2I=\log 2\left( x \right)_{0}^{\dfrac{\pi }{4}} \\
\end{align}$
On simplifying further we get,
$I=\dfrac{1}{2}\log 2\left( \dfrac{\pi }{4}-0 \right)$
$\begin{align}
  & I=\dfrac{{{\log }_{e}}2}{2}\times \dfrac{\pi }{4} \\
 & I=\dfrac{\pi }{8}{{\log }_{e}}2 \\
\end{align}$
Put value of $I$ from equation (1) we get,
$\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}=\dfrac{\pi }{8}{{\log }_{e}}2$

So, the correct answer is “Option A”.

Note: An integral assigns numbers to functions to describe the displacement, area, volume and other concepts. The process of finding the integrals is known as Integration. Integrals are of two types: indefinite and definite integral where indefinite integral doesn’t have any limit to which the integral is to be calculated whereas in definite integral the limit or summation is defined. Logarithm is an inverse function to exponential which means logarithm of any number $x$ is the exponent to which another fixed number $b$ is to be raised. Trigonometry is a branch of science which deals with the side lengths and angles of triangles. There are six trigonometric functions which are sine, cosine, tangent, cosecant, secant and cotangent.