How do you integrate #(tanx)^2#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Ratnaker Mehta Aug 21, 2016 #tanx-x+C#. Explanation: We will use the Trigo. Identity # : sec^2x=tan^2x+1#. Hence, #int(tanx)^2 dx=int tan^2xdx=int (sec^2x-1)dx# #=int sec^2xdx-int 1 dx=tanx-x+C#. Enjoy Maths.! 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 64161 views around the world You can reuse this answer Creative Commons License