How do you find the antiderivative of #cos^3 (x)dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Eddie Jun 23, 2016 # sin x- 1/3 sin^3x + C# Explanation: #\int cos^3 x \ dx = = \int cosx(cos^2x) \ dx= \int cosx(1-sin^2x) \ dx# and that's pretty much it because # \int cosx(1-sin^2x) \ dx# # = \int cosx- cosx sin^2x \ dx# # = sin x- 1/3 sin^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 11401 views around the world You can reuse this answer Creative Commons License