How do you find the antiderivative of #sin^2(x)cosx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Alexander Jul 10, 2016 #int sin^2(x) cos(x) dx#, let #u = sin(x) -> du = cos(x) dx# Thus #int u^2 du = 1/3 u^3 + C = (sin^3(x))/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 1397 views around the world You can reuse this answer Creative Commons License