What is the integration of sin inverse root x ?

Solution :

We have, I = \(sin^{-1}\sqrt{x}\) . 1 dx

By Applying integration by parts,

Taking \(sin^{-1}\sqrt{x}\) as first function and 1 as second function. Then

I = \(sin^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(sin^{-1}\sqrt{x}\) \(\int\) 1 dx } dx

I = x\(sin^{-1}\sqrt{x}\) – \(\int\) \(1\over 2\sqrt{(1-x)}\sqrt{x}\) . x dx

I = x\(sin^{-1}\sqrt{x}\) – \(\int\) \(1\over 2\) \(\sqrt{x}\over \sqrt{(1-x)}\) dx

Put x = \(sin^2 t\)

dx = 2 sin t cos t dt

\(\implies\) I = x\(sin^{-1}\sqrt{x}\) – { \(1\over 2\) \(\int\) \(2sin^2 t\) dt }

\(\implies\) I = x\(sin^{-1}\sqrt{x}\) – { \(1\over 2\) \(\int\) (1 – cos 2t) dt }

\(\implies\) I = x\(sin^{-1}\sqrt{x}\) – \(1\over 2\)t + \(1\over 4\)\(sin 2t\) + C

Now, sin 2t = 2 sin t cos t = \(2\sqrt{x} \sqrt{1 – x}\) = \(2\sqrt{x – x^2}\)

I = x\(sin^{-1}\sqrt{x}\) – \(1\over 2\)\(sin^{-1}\sqrt{x}\) + \(1\over 4\) (\(2\sqrt{x – x^2}\)) + C

I = (x – \({1\over 2}\))\(sin^{-1}\sqrt{x}\) + \(1\over 2\) \(\sqrt{x – x^2}\) + C