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 2661 views around the world You can reuse this answer Creative Commons License