Q185 / 270 · NDA 2010 (Phase 2) · Mathematics

What is the value of $\int_{\pi / 6}^{\pi / 4} \frac{d x}{\sin x \cos x} ?$

  • A
    $2 \ln \sqrt{3}$
  • Correct answer
    B
    $\ln \sqrt{3}$
  • C
    $2 \ln 3$
  • D
    $4 \ln 3$

Solution

$\begin{aligned} & \int_{\pi / 6}^{\pi / 4} \frac{-d x}{\sin x \cdot \cos x}=\int_{\pi / 6}^{\pi / 4} \frac{\sec ^2 x}{\tan x} d x \\ & =\int_{\frac{1}{\sqrt{3}}}^1 \frac{d z}{z}=[\log z]_{\frac{1}{\sqrt{3}}}^1 \\ & =[\log 1-(\log 1-\log \sqrt{3})]=\log \sqrt{3}\end{aligned}$

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

Relevant Questions