First slide

Evaluation of definite integrals

Question

$\int_{0}^{π / 4} \log (1 + \tan^{2} θ + 2 \tan θ) d θ =$

Easy

A

$π \log 2$
B

$(π \log 2) / 2$
C

$(π \log 2) / 4$
D

$\log 2$

Solution

Let $I = \int_{0}^{π / 4} \log (1 + \tan θ) d θ$
$\begin{array}{l} = \int_{0}^{π / 4} \log (1 + \tan (π / 4 - θ)) d θ \\ = \int_{0}^{π / 4} \log (1 + \frac{1 - \tan θ}{1 + \tan θ}) d θ \\ = \int_{0}^{π / 4} [\log 2 - \log (1 + \tan θ)] d θ \\ \Rightarrow 2 I = π / 4 \log 2 . \end{array}$
Required integral $= 2 I = \frac{π}{4} \log 2$

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