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 2429 views around the world You can reuse this answer Creative Commons License