How do you find the antiderivative of #int x^2/sqrt(4-x^2)dx#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Ratnaker Mehta Feb 22, 2017 # 2arc sin(x/2)-x/2sqrt(4-x^2)+C.# Explanation: We know that, # (1) : intsqrt(a^2-x^2)dx=x/2sqrt(a^2-x^2)+a^2/2arc sin(x/a)+c_1.# # (2) : int1/sqrt(a^2-x^2)dx=arc sin(x/a)+c_2.# Hence, #I=intx^2/sqrt(4-x^2)dx# #=-int(-x^2)/sqrt(4-x^2)dx=-int{(4-x^2)-4}/sqrt(4-x^2)dx,# #=-int(4-x^2)/sqrt(4-x^2)dx+4int1/sqrt(4-x^2)dx,# #=-intsqrt(4-x^2)dx+4arc sin(x/2),............[because, (2)]# #=-{x/2sqrt(4-x^2)+4/2arc sin(x/2)}+4arc sin (x/2),# #=2arc sin(x/2)-x/2sqrt(4-x^2)+C.# Enjoy Maths.! 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 1251 views around the world You can reuse this answer Creative Commons License