How do you find the integral of int (6sin^6x)(cos^3x)dx? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Sasha P. Oct 20, 2015 I = (6sin^7x)/7-(2sin^9x)/3+C Explanation: I = int (6sin^6x)(cos^3x)dx = 6int sin^6x cos^2x cosxdx I = 6int sin^6x (1-sin^2x) cosxdx sinx=t => cosxdx=dt I = 6int t^6 (1-t^2) dt = 6 int (t^6-t^8)dt = 6 (t^7/7-t^9/9+C) I = (6sin^7x)/7-(2sin^9x)/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 2603 views around the world You can reuse this answer Creative Commons License