What is the integration of x sin inverse x dx ?

Solution :

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

By using integration by parts formula,

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

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

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

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

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

By using integration formula of \(\sqrt{a^2 – x^2}\),

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

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


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