Integration Interval

let x=sec y so dx= sec y tan y dy...the limit of x also changes from 0 to pie/3...putting x= sec y and also the value of dx , the integration becomes easy...hence solve it.

$\begin{aligned} I & = \int_1^2 \frac 1{x\sqrt{x^2-1}} dx & \small \color{#3D99F6} \text{Let }x = \sec \theta \implies dx = \sec \theta \tan \theta \ d \theta \\ & = \int_0^\frac \pi 3 \frac {\sec \theta \tan \theta}{\sec \theta \sqrt{\sec^2 \theta -1}} d \theta & \small \color{#3D99F6} x = 1 \implies \theta = 0, \ x = 2 \implies \theta = \frac \pi 3 \\ & = \int_0^\frac \pi 3 \frac {\sec \theta \tan \theta}{\sec \theta \tan \theta} d \theta \\ & = \int_0^\frac \pi 3 d \theta = \theta \ \bigg|_0^\frac \pi 3 = \frac \pi 3 \end{aligned}$

$\implies n = \boxed{3}$

We Know That The Integration Of Above Function Is Sec inverse 1 Therefore by putting limit we get pie/3

From the identity of the derivative of the inverse tangent function, it can be seen that this integral equates to $\arctan{\sqrt{x^2-1}}$ .

On further inspection, this can be simplified to $-\arcsin{\frac{1} {|x|}}$ .

Substituting in the values given for the integral into $-\arcsin{\frac{1} {|x|}}$ , you get $-30 + 90$ , and converting this into radians, you get $\frac{\pi} {3}$ .

Therefore, n = 3

Q.E.D