Question

Integrate \[\log \left( 1+\tan x \right)\] from \[0\] to \[\dfrac{\pi }{4}\].

Answer 1

Hint: In order to integrate the given function \[\log \left( 1+\tan x \right)\] from \[0\] to \[\dfrac{\pi }{4}\], we will be considering the given limits by substituting them in the function given. Then by applying the suitable integration formulas, we will be solving the given function and obtaining the required solution.

Complete step-by-step solution:
Now let us have a brief regarding integration. It is nothing but calculating the integral. The integrals are usually termed regarding the definite integrals and indefinite integrals are used for antiderivatives. Integration is the reverse process of differentiation. There are two types of integrals. They are: definite and indefinite integrals. Integration can be performed in different methods. They are: Integration by Substitution, Integration by Parts, Integration using Trigonometric Identities, Integration of Some Particular Function and Integration by Partial Fraction.
Now let us perform integration on \[\log \left( 1+\tan x \right)\] from \[0\] to \[\dfrac{\pi }{4}\].
Let us write it as \[I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)}\to \left( 1 \right)\]
Using the formula \[\int\limits_{0}^{a}{f\left( x \right)dx=\int\limits_{0}^{a}{f\left( a-x \right)dx}}\], let us express the given function. We get
\[\Rightarrow \]\[I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ 1+\tan \left( \dfrac{\pi }{4}-x \right) \right]dx}\]
Upon solving this,
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ 1+\dfrac{\tan \dfrac{\pi }{4}-\tan x}{1+\tan \dfrac{\pi }{4}\tan x} \right]}dx\] \[\left[ \because \left( \tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b} \right) \right]\]
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ \dfrac{1-\tan x}{1+1\cdot \tan x} \right]}dx\] We know that \[\left( \tan \dfrac{\pi }{4}=1 \right)\]
Then solving accordingly, we obtain
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ \dfrac{1-\tan x+1-\tan x}{1+\tan x} \right]}dx\]
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ \dfrac{2}{1+\tan x} \right]}dx\]
\[\Rightarrow I=\]\[\int\limits_{0}^{\dfrac{\pi }{4}}{\left[ \log 2-\log \left( 1+\tan x \right) \right]}\] \[\left[ \because \log \dfrac{a}{b}=\log a-lobg \right]\]
Splitting up according to the formula, we get
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx-\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx\to \left( 2 \right)}}\]
Now upon adding the two equations, we get
\[\begin{align}
  & I+I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)+}\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx-\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}} \\
 & 2I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx} \\
 & \Rightarrow 2I=\log 2\int\limits_{0}^{\dfrac{\pi }{4}}{dx} \\
 & \Rightarrow I=\dfrac{\log 2}{2}\left[ x \right]_{0}^{\dfrac{\pi }{4}} \\
 & \Rightarrow I=\dfrac{\log 2}{2}\left[ \dfrac{\pi }{4}-0 \right] \\
 & \Rightarrow I=\dfrac{\log 2}{2}\times \dfrac{\pi }{4} \\
 & \Rightarrow I=\dfrac{\pi }{8}\log 2 \\
\end{align}\]

Note: While performing the integration, we must be aware of the sum rule, power rule, difference rule, etc. While solving, we must cross check with the integration rules for easy solving of the functions given. We can apply integration in finding the areas, volumes, central points and many more.