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Question

Mathematics Question on integral

Integrate the function: xsin−1xxsin^{-1}x

Answer

LetLet II=∫$$xsin^{-1}x\ dx

Taking as first function and x as second function and integrating by parts, we obtain

I= sin−1x∫[xdx−∫(ddxsin−1x)∫xdx]dxsin^{-1}x∫[x dx-∫{(\frac {d}{dx}sin^{-1}x)}∫x dx]dx

I=sin−1x(x22)−∫11−x2.x22dxI= sin^{-1}x (\frac {x^2}{2})-∫\frac {1}{\sqrt {1-x^2}}.\frac {x^2}{2} dx

I=x2sin−1x2+12∫−x21−x2dxI= \frac {x^2sin^{-1}x}{2}+\frac 12∫\frac {-x^2}{\sqrt {1-x^2}} dx

I= x2sin−1x2\frac {x^2sin^{-1}x}{2} + 12∫[1−x21−x2−1√1−x2]dx\frac 12∫[{\frac {1-x^2}{\sqrt {1-x^2}}-\frac {1}{√1-x^2}}]dx

I= x2sin−1x2\frac {x^2sin^{-1}x}{2} + 12∫[1−x2−11−x2]dx\frac 12∫[{\sqrt {1-x^2}-\frac {1}{\sqrt {1-x^2}}}]dx

I= x2sin−1x2\frac {x^2sin^{-1}x}{2} + 12∫x21−x2 dx−∫11−x2 dx\frac 12∫\frac x2{\sqrt {1-x^2\ }dx-∫\frac {1}{\sqrt {1-x^2}}}\ dx

I= x2sin−1x2\frac {x^2sin^{-1}x}{2} + 12[x21−x2+12sin−1x−sin−1x]+C\frac 12[{\frac x2\sqrt {1-x^2}+\frac 12sin^{-1}x-sin^{-1}x}]+C

I= x2sin−1x2\frac {x^2sin^{-1}x}{2} + x41−x2+14sin−1x−12sin−1x+C{\frac x4\sqrt {1-x^2}+\frac 14sin^{-1}x-\frac 12sin^{-1}x}+C

I=14(2x2−1)sin−1x+x41−x2+CI= \frac 14(2x^2-1)sin^{-1}x+\frac x4\sqrt {1-x^2}+C