How do you find #int 1-tan^2xdx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Jim H · Tom Nov 15, 2015 Use trigonometric identity: #tan^2x = sec^2x-1# to rewrite. Explanation: #int (1-tan^2x)dx = int(1-(sec^2x-1))dx# # = int(2-sec^2x) dx# # = 2x-tanx +C# 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 4423 views around the world You can reuse this answer Creative Commons License