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
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