Question

Evaluate: ${\int }_1^2\frac{1}{x\sqrt{x^2-1}}dx$

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (D)

$\frac{\pi }{3}$

We need to find value of ${\int }_1^2\frac{1}{x\sqrt{x^2-1}}dx$
We know that

$\int \frac{1}{x\sqrt{x^2-1}}dx={\sec }^{-1}x+c$

${\int }_1^2\frac{1}{x\sqrt{x^2-1}}dx={\sec }^{-1}2-{\sec }^{-1}1$

${\sec }^{-1}2=\frac{\pi }{3}$

and ${\sec }^{-1}1=0$

So, ${\sec }^{-1}2-{\sec }^{-1}1=\frac{\pi }{3}$

$\therefore {\int }_1^2\frac{1}{x\sqrt{x^2-1}}dx=\frac{\pi }{3}$