Integration Questions

What is the integration of cos inverse root x ?

Solution : We have, I = \(cos^{-1}\sqrt{x}\) . 1 dx By Applying integration by parts, Taking \(cos^{-1}\sqrt{x}\) as first function and 1 as second function. Then I = \(cos^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(cos^{-1}\sqrt{x}\) \(\int\) 1 dx } dx I = x\(cos^{-1}\sqrt{x}\) – \(\int\) \(-1\over 2\sqrt{(1-x)}\sqrt{x}\) . x dx I = x\(cos^{-1}\sqrt{x}\) – …

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What is the integration of x cos inverse x ?

Solution : We have, I = \(\int\)  \(x cos^{-1} x\) dx By using integration by parts formula, I = \(cos^{-1} x\) \(x^2\over 2\) – \(\int\) \(-1\over \sqrt{1 – x^2}\) \(\times\) \(x^2\over 2\) dx I =  \(x^2\over 2\) \(cos^{-1} x\) – \(1\over 2\) \(\int\) \(-x^2\over \sqrt{1 – x^2}\) dx = \(x^2\over 2\) \(cos^{-1} x\) – \(1\over …

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