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MHT CET · Maths · Definite Integration

\(\int_0^{\pi / 4} \log (1+\tan x) d x=\)

  1. A \(\frac{\pi}{16} \log 2\)
  2. B \(\frac{\pi}{4} \log 2\)
  3. C \(\frac{\pi}{8} \log 2\)
  4. D \(\pi \log 2\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{\pi}{8} \log 2\)

Step-by-step Solution

Detailed explanation

\( \text { Let } I=\int_0^{\pi / 4} \log (1+\tan x) d x \)
\( =\int_0^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x \)
\( =\int_0^{\pi / 4} \log \left[1+\left(\frac{1-\tan x}{1+\tan x}\right)\right] d x=\int_0^{\pi / 4}\) \(\log \left(\frac{2}{1+\tan x}\right) d x \)
\( =\int_0^{\pi / 4}(\log 2) d x-\int_0^{\pi / 4} \log (1+\tan x) d x=\) \(\int_0^{\pi / 4}(\log 2) d x-I \)
\( \therefore 2 I=(\log 2)[x]_0^{\pi / 4}=(\log 2)\left(\frac{\pi}{4}\right) \Rightarrow I=\left(\frac{\pi}{8}\right) \log 2\)