How do you find the integral of #cos^3 x#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Jim H Oct 20, 2015 Use #cos^3x = cos^2x cosx = (1-sin^2x) cosx = cosx - sin^2xcosx# Explanation: So, #int cos^3x dx = int cosx dx - int sin^2xcosx dx# #=sinx -sin^3x/3 +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 1076 views around the world You can reuse this answer Creative Commons License