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How do you integrate #int sin^-1x# by integration by parts method?

1 Answer

Apr 8, 2018

#xarcsinx+sqrt(1-x^2)+C#

Explanation:

After choosing #u=arcsinx# and #dv=dx#, #du=(dx)/sqrt(1-x^2)# and #v=x#

Hence,

#int arcsinx*dx=xarcsinx-int x*(dx)/sqrt(1-x^2)#

=#xarcsinx-int (xdx)/sqrt(1-x^2)#

=#xarcsinx+sqrt(1-x^2)+C#

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