How do you find the antiderivative of #(cos(3x))^3#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer sjc Oct 22, 2017 #1/3sin3x-1/9sin^3 3x+c# Explanation: #intcos^3 3xdx# #=intcos3xcos^2 3xdx# #=intcos3x(1-sin^2 3x)dx# #I=int(cos3x-cos3xsin^2 3x)dx# now #d/(dx)(sin^3 3x)=3 xx3xxsin^2 3xcos3x=9sin^2 3xcos3x# #I=1/3sin3x-1/9sin^3 3x+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 1952 views around the world You can reuse this answer Creative Commons License