How do you find the antiderivative of #(2x)(sinx)(cosx)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer ali ergin May 29, 2016 #int 2x*sin x*cos x d x# #int x*2sin x cos x d x# #2sin x cos x=sin 2x# #int x* sin 2x *d x=-1/2x*cos 2x+1/4*sin 2x+C# #int2x*sin x*cos x d x=-1/2x*cos 2x+1/4*sin 2x+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 1169 views around the world You can reuse this answer Creative Commons License