
Find the value of $\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan xdx}$.
A. $\dfrac{\pi }{2}{{\log }_{e}}2$
B. $-\dfrac{\pi }{2}{{\log }_{e}}2$
C. $\pi {{\log }_{e}}2$
D. $0$
Answer
162.3k+ views
Hint: In this question, we have $\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan xdx}$. So, we will consider this equation as 1. Now we will use the identity $\int_{0}^{a}{f\left( x \right)dx=}\int_{0}^{a}{f\left( a-x \right)dx}$. After that, we will get equation 2.
Now we will add equations 1 and 2 and then we will use the identity $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$. At last, we will obtain the final result.
Formula Used:For the given problem, we will use the following formulas:
1) $\int_{0}^{a}{f\left( x \right)dx=}\int_{0}^{a}{f\left( a-x \right)dx}$
2) $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$
3) $\log \left( 1 \right)=0$
Complete step by step solution:We are given an integral as $\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan xdx}$.
Let us consider $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan xdx}$ ….... (1)
First, we will use the identity $\int_{0}^{a}{f\left( x \right)dx=}\int_{0}^{a}{f\left( a-x \right)dx}$.
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan \left( \dfrac{\pi }{2}-x \right)}dx$
As we know that $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$
So, the integral becomes
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \cot xdx}$ ….... (2)
Now, we are adding equations (1) and (2).
$\begin{align}
& \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan xdx+\int\limits_{0}^{\dfrac{\pi }{2}}{\log \cot xdx}} \\
& \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log (\tan x\times \cot x)dx} \\
\end{align}$
Further, we will write $\cot x$ as the reciprocal of $\tan x$. So, we get
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \left( \tan x\times \dfrac{1}{\tan x} \right)}dx$
By canceling the same terms, we get
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log (1)dx}$
Remember that the value of $\log \left( 1 \right)=0$.
$\begin{align}
& \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{(0)dx} \\
& \Rightarrow 2I=0 \\
& \therefore I=0 \\
\end{align}$
Option ‘D’ is correct
Note: As the functions change, it is not essential to integrate taking limits directly $0$ to $\dfrac{\pi }{2}$. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier. Students are advised to write the correct identity and then solve the problem. Don’t choose the identity that makes the question more complicated.
Now we will add equations 1 and 2 and then we will use the identity $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$. At last, we will obtain the final result.
Formula Used:For the given problem, we will use the following formulas:
1) $\int_{0}^{a}{f\left( x \right)dx=}\int_{0}^{a}{f\left( a-x \right)dx}$
2) $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$
3) $\log \left( 1 \right)=0$
Complete step by step solution:We are given an integral as $\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan xdx}$.
Let us consider $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan xdx}$ ….... (1)
First, we will use the identity $\int_{0}^{a}{f\left( x \right)dx=}\int_{0}^{a}{f\left( a-x \right)dx}$.
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan \left( \dfrac{\pi }{2}-x \right)}dx$
As we know that $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$
So, the integral becomes
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \cot xdx}$ ….... (2)
Now, we are adding equations (1) and (2).
$\begin{align}
& \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \tan xdx+\int\limits_{0}^{\dfrac{\pi }{2}}{\log \cot xdx}} \\
& \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log (\tan x\times \cot x)dx} \\
\end{align}$
Further, we will write $\cot x$ as the reciprocal of $\tan x$. So, we get
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \left( \tan x\times \dfrac{1}{\tan x} \right)}dx$
By canceling the same terms, we get
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log (1)dx}$
Remember that the value of $\log \left( 1 \right)=0$.
$\begin{align}
& \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{(0)dx} \\
& \Rightarrow 2I=0 \\
& \therefore I=0 \\
\end{align}$
Option ‘D’ is correct
Note: As the functions change, it is not essential to integrate taking limits directly $0$ to $\dfrac{\pi }{2}$. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier. Students are advised to write the correct identity and then solve the problem. Don’t choose the identity that makes the question more complicated.
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Advanced 2025 Notes
