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 1908 views around the world You can reuse this answer Creative Commons License