What is the integral of #tan^2(x)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Mika Dec 19, 2014 Use the identity #tan^2x=sec^2x-1# #int tan^2x dx# #=int sec^2x - 1 dx# #=int sec^2x dx - int dx# Integrate separately and you'll get. #tanx - x# Answer link Related questions How do I evaluate the indefinite integral #intsin^3(x)*cos^2(x)dx# ? How do I evaluate the indefinite integral #intsin^6(x)*cos^3(x)dx# ? How do I evaluate the indefinite integral #intcos^5(x)dx# ? How do I evaluate the indefinite integral #intsin^2(2t)dt# ? How do I evaluate the indefinite integral #int(1+cos(x))^2dx# ? How do I evaluate the indefinite integral #intsec^2(x)*tan(x)dx# ? How do I evaluate the indefinite integral #intcot^5(x)*sin^4(x)dx# ? How do I evaluate the indefinite integral #inttan^2(x)dx# ? How do I evaluate the indefinite integral #int(tan^2(x)+tan^4(x))^2dx# ? How do I evaluate the indefinite integral #intx*sin(x)*tan(x)dx# ? See all questions in Integrals of Trigonometric Functions Impact of this question 1716 views around the world You can reuse this answer Creative Commons License