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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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Best answer
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,

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

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

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