How do you integrate #int cot^2xdx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer sjc Dec 1, 2016 #-cotx-x+C# Explanation: the identity#" "1+cot^2x=csc^2x" "#is used. #cot^2x=csc^2x-1# #intcot^2xdx=int(csc^2x-1)dx# #=intcsc^2xdx-intdx# #=-cotx-x+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 73098 views around the world You can reuse this answer Creative Commons License