Q269 / 270 · NDA 2021 (Phase 2) · Mathematics

What is $\int_0^\pi \ln \left(\tan \frac{x}{2}\right) d x$ equal to?

  • Correct answer
    A
    0
  • B
    $\frac{1}{2}$
  • C
    1
  • D
    2

Solution

$\begin{aligned} \mathrm{I} & =\int_0^\pi \ln \left(\tan \frac{x}{2}\right) d x \qquad \qquad...\mathrm{(i)}\\ \mathrm{I} & =\int_0^\pi \ln \left(\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right) d x \\ & =\int_0^\pi \ln \left(\cot \frac{x}{2}\right) d x \qquad \qquad...\mathrm{(ii)} \end{aligned}$ Adding equation (1) and (2), we get $2 \mathrm{I}=\int_0^\pi \ln \left(\tan \frac{x}{2}\right) d x+\int_0^\pi \ln \left(\cot \frac{x}{2}\right) d x$ $\begin{aligned} 2 \mathrm{I} & =\int_0^\pi \ln \left(\tan \frac{x}{2} \cot \frac{x}{2}\right) d x \\ 2 \mathrm{I} & =\int_0^\pi \ln \left(\tan \frac{x}{2} \times \frac{1}{\tan \frac{x}{2}}\right) d x \\ \mathrm{I} & =\frac{1}{2} \int_0^\pi \ln (1) d x \quad(\therefore \ln 1=0) \\ & \therefore \mathrm{I}=0\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions