How do you integrate #int tan^2x#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer sjc Nov 25, 2016 #int tan^2xdx=tanx-x+C# Explanation: use the identity #" "1+tan^2x=sec^2x# #=>tan^2x=sec^2x-1# #:.inttan^2xdx=int(sec^2x-1)dx# #=tanx-x+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 1090 views around the world You can reuse this answer Creative Commons License