What is the antiderivative of # (2x)(sinx)(cosx)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Konstantinos Michailidis May 27, 2016 We have that #int (2x)(sinx)(cosx)dx=int x*sin2xdx= -1/2 int x (cos2x)'dx=-1/2*[xcos2x-intcos2xdx]= -1/2xcos2x+1/4*sin2x+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 1708 views around the world You can reuse this answer Creative Commons License