How do you integrate #int sin^-1x# by integration by parts method? Calculus Techniques of Integration Integration by Parts 1 Answer Cem Sentin 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# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 42929 views around the world You can reuse this answer Creative Commons License