Question

integration of Sin inverse X

Answer 1

hey friend here is your answer^^ (I have also attached soft copy with this ans for your better understanding)

∫sin−1xdx∫sin−1⁡xdx

Integrating by parts by taking sin−1xsin−1⁡x as the first function

∫sin−1x.1dx∫sin−1⁡x.1dx

sin−1x∫1dx−∫(ddx(sin−1x)∫1dx)dxsin−1⁡x∫1dx−∫(ddx(sin−1⁡x)∫1dx)dx

xsin−1x−∫x1−x2−−−−−√dxxsin−1⁡x−∫x1−x2dx

1−x2=t1−x2=t

−2xdx=dt−2xdx=dt

xdx=−12dtxdx=−12dt

xsin−1x+12∫dtt√xsin−1⁡x+12∫dtt

xsin−1x+122t√+Cxsin−1⁡x+122t+C

xsin−1x+t√+Cxsin−1⁡x+t+C

xsin−1x+1−x2−−−−−√+Cxsin−1⁡x+1−x2+C