How do you integrate #int ((4-x)^2)^(3/2)dx# using trigonometric substitution? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer James May 1, 2018 The answer #-[1/4*(4-x)^4]+c# Explanation: Show the steps #int ((4-x)^2)^(3/2)dx=int(4-x)^(3/2*2)*dx=int(4-x)^3*dx# #-[1/4*(4-x)^4]+c# Note that #((x)^n)^m=x^(n*m)# 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 6590 views around the world You can reuse this answer Creative Commons License