# What is $\rm \displaystyle\int_0^1 \dfrac{\sin^{-1} \ x}{\sqrt {{1-x^2}}} \ dx$ equal to?

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What is $\rm \displaystyle\int_0^1 \dfrac{\sin^{-1} \ x}{\sqrt {{1-x^2}}} \ dx$ equal to?
1. $\rm \dfrac{\pi^2}{2}$
2. $\rm \dfrac{\pi^2}{4}$
3. $\rm \dfrac{\pi^2}{8}$
4. $\rm \dfrac{\pi^2}{16}$

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Correct Answer - Option 3 : $\rm \dfrac{\pi^2}{8}$

Concept:

$\rm \displaystyle\int x^n \;dx = \dfrac{x^{n+1}}{n+1} + c$

$\rm {d(sin^{-1}x)\over dx} = {1\over {\sqrt {1\ -\ x^2}}}$

Calculation:

Given: $\rm \displaystyle\int_0^1 \dfrac{\sin^{-1} \ x}{\sqrt {{1-x^2}}} \ dx$

Let sin-1 x = t

$\Rightarrow \rm \frac{dx}{\sqrt {1-x^2}} = dt$

 x 0 1 t 0 π/2

Now,

$\rm \rm \displaystyle\int_0^1 \dfrac{sin^{-1} \ x}{\sqrt {{1-x^2}}} \ dx = \rm \displaystyle\int_{0}^{\pi /2} t\;dt$

$\rm = \left[\dfrac{t^2}{2}\right]_{0}^{\pi/2} \\ = \dfrac{\pi^2}{8}$