Integration Questions

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 […]

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\(sin^{-1}\sqrt{x}\) –

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 = \(t^2\)

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

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

Solution : Let I = \(\int\) x\(tan^{-1}x\) dx By using Integration by parts rule, Taking tan inverse x as first function and x as second function. Then, I = (\(tan^{-1}x\)) \(\int\) x dx – \(\int\){\({d\over dx}\)(\(tan^{-1}x\) \(\int\) x dx} dx I = (\(tan^{-1}x\))\(x^2\over 2\) – \(\int\)\({1\over 1 + x^2}\) \(\times\) \(x^2\over 2\) dx \(\implies\) I

Solution : Let I = \(\int_{0}^{\pi/2}\) log(sinx)dx    …….(i) then I = \(\int_{0}^{\pi/2}\) \(log(sin({\pi\over 2}-x))\)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx     …….(ii) Adding (i) and (ii), we get 2I = \(\int_{0}^{\pi/2}\) log(sinx)dx + \(\int_{0}^{\pi/2}\) log(cosx)dx = \(\int_{0}^{\pi/2}\) (log(sinx)dx + log(cosx)) \(\implies\) \(\int_{0}^{\pi/2}\) log(sinxcosx)dx = \(\int_{0}^{\pi/2}\) \(log({2sinxcosx\over 2})\)dx = \(\int_{0}^{\pi/2}\) \(log({sin2x\over 2})\)dx = \(\int_{0}^{\pi/2}\) log(sin2x)dx – \(\int_{0}^{\pi/2}\) log(2)dx

Solution : I = \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\) = \(\int\) \(cos^4xdx\over {sin^6x{(1 + cot^5x)^{3\over 5}}}\) = \(\int\) \(cot^4xcosec^2xdx\over {{(1 + cot^5x)^{3\over 5}}}\) Put \(1+cot^5x\) = t \(5cot^4xcosec^2x\)dx = -dt = -\(1\over 5\) \(\int\) \(dt\over {t^{3/5}}\) = -\(1\over 2\) \(t^{2/5}\) + C = -\(1\over 2\) \({(1+cot^5x)}^{2/5}\) + C Similar Questions What is the integration

Solution : I = \(\int\) \(dx\over {3sinx + 4cosx}\) = \(\int\) \(dx\over {3[{2tan{x\over 2}\over {1+tan^2{x\over 2}}}] + 4[{1-tan^2{x\over 2}\over {1+tan^2{x\over 2}}}]}\) = \(\int\) \(sec^2{x\over 2}dx\over {4+6tan{x\over 2}-4tan^2{x\over 2}}\) let \(tan{x\over 2}\) = t, \(\therefore\)  \({1\over 2}sec^2{x\over 2}\)dx = dt so I = \(\int\) \(2dt\over {4+6t-4t^2}\) = \(1\over 2\) \(\int\) \(dt\over {1-(t^2-{3\over 2}t})\) = \(1\over 2\)