Mathematics

# Evaluate: $\int _{ 0 }^{ \pi /4 }{ \log { \left( 1+\tan { \theta } \right) } } d\theta$

##### SOLUTION
Let $\displaystyle I=\int_{0}^{\dfrac{\pi }{4}}\ln(1+ \tan \theta )d\theta$

using $\displaystyle \int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$

$I=\displaystyle \int_{\dfrac{\pi }{4}}^{0}\ln\left(1+ \tan \left(\dfrac{\pi }{4}-\theta \right)\right)d\theta$

$I= \displaystyle \int_0^{\dfrac{\pi}{4}}\ln \left( 1 + \dfrac{1-\tan \theta}{1+\tan \theta}\right) d \theta$

$I=\displaystyle \int_{0}^{\dfrac{\pi }{4}}\ln\left(\dfrac{1+ \tan \theta +1- \tan \theta }{1+ \tan \theta }\right)d\theta$

$I=\displaystyle \int_{0}^{\dfrac{\pi }{4}}(\ln (2)-\ln(1+ \tan \theta))d\theta$

$I=\displaystyle \int_{0}^{\dfrac{\pi }{4}}\ln2\,d\theta-I$

$2I=\dfrac{\pi }{4}\ln2\Rightarrow I=\dfrac{\pi }{8}\ln 2$

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

#### Realted Questions

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lf $\displaystyle \int_{0}^{1}\frac{\tan^{-1}x}{x}dx=k\int_{0}^{\pi/2}\frac{x}{\sin x}dx$, then the value of $k$ is
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• B. $\displaystyle \frac{1}{4}$
• C. $4$
• D. ${2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
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