How do you find the antiderivative of #(cosx)^2#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Shwetank Mauria Jul 27, 2016 #intcos^2xdx=x/2+(sin2x)/4+c# Explanation: To find antiderivative i.e. integral of #cos^2x#, we can use formula #cos^2x=1/2(1+cos2x)# #intcos^2xdx=int[1/2(1+cos2x)]dx# = #int(1/2+(cos2x)/2)dx# = #1/2[x+(sin2x)/2]+c# = #x/2+(sin2x)/4+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 17705 views around the world You can reuse this answer Creative Commons License