How do you find the antiderivative of #Cos(2x)Sin(x)dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Monzur R. Aug 30, 2017 #intcos2xsinxdx = cosx-2/3cos^3x + "constant"# Explanation: # cos2x -= 2cos^2x-1# #therefore intcos2xsinxdx = int(2cos^2x -1)sinxdx = 2intcos^2xsinxdx -intsinxdx# Let #u = cosx # and #du = -sinxdx# Then #cosx +2intcos^2xsinxdx = cosx-2intu^2du = cosx-2/3u^3=cosx-2/3cos^3x +"constant"# 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 1443 views around the world You can reuse this answer Creative Commons License