Indefinite Integration 1 Question 71
72. Integrate $\int _0^{\pi / 4} \log (1+\tan x) d x$.
(1997C, 2M)
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Solution:
- Let $I=\int _0^{\pi / 4} \log (1+\tan x) d x$
$$ \begin{aligned} & I=\int _0^{\pi / 4} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\ \therefore \quad & I=\int _0^{\pi / 4} \log 1+\frac{1-\tan x}{1+\tan x} d x \end{aligned} $$
$$ \begin{aligned} & \qquad \int _0^{\pi / 4} \log \frac{1+\tan x+1-\tan x}{1+\tan x} d x \\ & \qquad I=\int _0^{\pi / 4} \log \frac{2}{1+\tan x} d x \Rightarrow I=\int _0^{\pi / 4} \log 2 d x-I \\ & \Rightarrow \quad 2 I=\frac{\pi}{4} \log 2 \Rightarrow \quad I=\frac{\pi}{8}(\log 2) \\ & \text {
