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Question

Mathematics Question on integral

∫sin−1xdx\int sin^{-1} x dx is equal to

A

cos−1x+Ccos^{- 1 }x + C

B

xsin−1x+1−x2+C xsin^{-1} x+\sqrt{1-x^{2} }+C

C

11−x2+C \frac{1}{\sqrt{1-x^{2}}}+C

D

xsin−1x−1−x2+C x sin^{-1}x-\sqrt{1-x^{2}}+C

Answer

xsin−1x+1−x2+C xsin^{-1} x+\sqrt{1-x^{2} }+C

Explanation

Solution

Let I=∫sin−1x⋅1dx=(sin−1x)x−∫11−x2⋅xdx+C′I=\int sin^{-1}x \cdot 1 dx = \left(sin^{-1}x\right)x -\int\frac{1}{\sqrt{1-x^{2}}}\cdot x dx +C' Put1−x2=t2⇒−2xdx=2tdtPut 1-x^{2} =t^{2 }\Rightarrow -2x dx = 2t dt xsin−1x−∫(−tdt)t+C′=xsin−1x+1−x2+Cx sin^{-1}x - \int\frac{\left(-t dt \right)}{t}+ C' = x sin^{-1} x + \sqrt{1-x^{2}}+C