How do you find the antiderivative of #cos^3 x#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Euan S. Jul 5, 2016 #int cos^3x dx = sinx - 1/3sin^3x + C# Explanation: #cos^3x = cosx*cos^2x = cosx(1 - sin^2x)# #implies int cos^3xdx = int cosx(1 - sin^2x)dx# Let #u = sinx implies du = cosxdx# #int (1-u^2)du = u - 1/3u^3 + C = sinx - 1/3sin^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 1242 views around the world You can reuse this answer Creative Commons License