Solution :
We have, I = \(\int\)ย \(x cos^{-1} x\) dx
By using integration by parts formula,
I = \(cos^{-1} x\) \(x^2\over 2\) โ \(\int\) \(-1\over \sqrt{1 โ x^2}\) \(\times\) \(x^2\over 2\) dx
I =ย \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) \(\int\) \(-x^2\over \sqrt{1 โ x^2}\) dx
= \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) \(\int\) \(1 โ x^2 โ 1\over \sqrt{1 โ x^2}\)ย dx
= \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) { \(\int\) \(1 โ x^2\over \sqrt{1 โ x^2}\) โ \(\int\) \(1\over \sqrt{1 -x^2}\) } dx
\(\implies\) I = \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) { \(\int\) \(\sqrt{1 โ x^2}\) โ \(\int\) \(1\over \sqrt{1 -x^2}\) } dx
By using integration formula of \(\sqrt{a^2 โ x^2}\),
\(\implies\) I = \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) [{ \(1\over 2\) \(x\sqrt{1 โ x^2}\) โ \(1\over 2\) \(sin^{-1} x\) } โ \(sin ^{-1} x\) ] + C
\(\implies\) I = \(x^2\over 2\) \(cos^{-1} x\) โย \(1\over 4\) \(x\sqrt{1 โ x^2}\) + \(3\over 4\) \(sin^{-1} x\) + C
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