What is #int ((arcsinx)^9) / (sqrt(1-x^2) dx#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer James May 29, 2018 #color(blue)[int ((arcsinx)^9) / (sqrt(1-x^2)] dx=(arcsinx)^10/10+c]# Explanation: #int ((arcsinx)^9) / (sqrt(1-x^2)] dx# lets suppose: #u=arcsinx# #du=1/sqrt(1-x^2)*dx# #dx=sqrt(1-x^2)*du# #int ((arcsinx)^9) / (sqrt(1-x^2) dx]=int ((u)^9*sqrt(1-x^2))/ (sqrt(1-x^2)]*du# #intu^9*du=u^10/10=(arcsinx)^10/10+c# Answer link Related questions How do you find the integral #int1/(x^2*sqrt(x^2-9))dx# ? How do you find the integral #intx^3/(sqrt(x^2+9))dx# ? How do you find the integral #intx^3*sqrt(9-x^2)dx# ? How do you find the integral #intx^3/(sqrt(16-x^2))dx# ? How do you find the integral #intsqrt(x^2-1)/xdx# ? How do you find the integral #intsqrt(x^2-9)/x^3dx# ? How do you find the integral #intx/(sqrt(x^2+x+1))dx# ? How do you find the integral #intdt/(sqrt(t^2-6t+13))# ? How do you find the integral #intx*sqrt(1-x^4)dx# ? How do you prove the integral formula #intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C# ? See all questions in Integration by Trigonometric Substitution Impact of this question 2318 views around the world You can reuse this answer Creative Commons License