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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\(cos^{-1}\sqrt{x}\) – …

What is the integration of cos inverse root x ? Read More »

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 …

What is the integration of x cos inverse x ? Read More »

Solution : We have, I = \(\int\) \(sin^{-1}(cos x)\) dx By using integration formula, cos x = \(sin({\pi\over 2} – x)\) I = \(\int\) \(sin^{-1}[sin({\pi\over 2} – x)]\) dx I = \(\int\) \(({\pi\over 2} – x)\) dx I = \({\pi\over 2}x\) – \(x^2\over 2\) + C Similar Questions What is the integration of sin inverse …

What is integration of sin inverse cos x ? Read More »

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\(sin^{-1}\sqrt{x}\) – …

What is the integration of sin inverse root x ? Read More »

Solution : We have, I = \((sin^{-1}x)^2\) dx Let \(sin^{-1}x\) = t, Then, x = sin t \(\implies\) dx = cos t dt \(\therefore\) I = \(\int\) \((sin^{-1}x)^2\) dx I = \(\int\) \(t^2\) cos t dt Applying integration by parts and, Taking \(t^2\) as first function and cos t as second function, I = \(t^2\) …

What is the integration of sin inverse x whole square ? Read More »

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 …

What is the integration of x sin inverse x dx ? Read More »