How do you find the integral of sin^3(x) cos^2(x) dx? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Sasha P. Oct 12, 2015 I = 1/5cos^5x-1/3cos^3x+C Explanation: I = int sin^3xcos^2xdx = int sin^2xcos^2xsinxdx I = int (1-cos^2x)cos^2xsinxdx cosx=t => -sinxdx=dt => sinxdx=-dt I = int (1-t^2)t^2(-dt) = int (t^4-t^2)dt = t^5/5-t^3/3+C I = 1/5cos^5x-1/3cos^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 84426 views around the world You can reuse this answer Creative Commons License