Mathematics

# Evaluate: $\displaystyle \int \frac{\ln (\tan x)}{\sin x\cos x}dx$

$\displaystyle \frac{1}{2}(\ln (\tan x))^{2}+c$

##### SOLUTION
$\displaystyle I=\int \frac{\ln (\tan x)}{\sin x\:\cos x}dx$

Let $\displaystyle t=\ln (\tan x)\Rightarrow dt=\dfrac{\sec ^{2}x}{\tan x}dx=\dfrac{dx}{\sin x\:\cos x}$

$\therefore \displaystyle I=\int tdt=\frac{1}{2}t^{2}+C$

$\displaystyle I=\frac{1}{2}(\ln (\tan x))^{2}+C$

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
The value of $\displaystyle\int\limits_{1}^{e^2}\dfrac{dx}{x}$
• A. $1$
• B. $-1$
• C. $-2$
• D. $2$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Integrate : $\displaystyle \int\sqrt{ax+b}dx$
• A. $\dfrac {3(ax+b)^{3/2}}{2a}+C$
• B. $\dfrac {1}{2\sqrt{ax+b}}+C$
• C. None of these
• D. $\dfrac {2(ax+b)^{3/2}}{3a}+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int \dfrac {dx}{(x^{2} + 4x + 5)^{2}}$ is equal to
• A. $\dfrac {1}{2}\left [\tan^{-1}(x + 1) + \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
• B. $\dfrac {1}{2}\left [\tan^{-1}(x + 2) - \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
• C. $\dfrac {1}{2}\left [\tan^{-1}(x + 1) - \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
• D. $\dfrac {1}{2}\left [\tan^{-1}(x + 2) + \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the definite integral :
$\displaystyle \int_{0}^{2}\dfrac {1}{4+x-x^2}dx$

Integrate the function    $x\sin^{-1}x$