How do you find the antiderivative of sin^3(x) cos^2(x) dxsin3(x)cos2(x)dx? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Ultrilliam May 3, 2018 = - 1/3 cos^3 x + 1/5cos^5 x + C=−13cos3x+15cos5x+C Explanation: int \ sin^3 x \ cos^2 x \ dx int \ sin x( 1- cos^2 x) \ cos^2 x \ dx = int \ sin x \ cos^2 x - sin x cos^4 x \ dx = int \ ( - 1/3 cos^3 x)^' - ( -1/5cos^5 x)^' \ dx = - 1/3 cos^3 x + 1/5cos^5 x + 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 1171 views around the world You can reuse this answer Creative Commons License