How do you integrate #cos^3 x#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Daniel L. Nov 28, 2016 #intcos^3xdx=-sin^3x/3+sinx+C# See explanation. Explanation: #intcos^3xdx=intcos^2x*cosxdx=int(1-sin^2x)cosxdx=|(t=sinx),(dt=cosxdx)|=int1-t^2dt=t-t^3/3+C = -sin^3x/3+sinx+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 2289 views around the world You can reuse this answer Creative Commons License