What is the antiderivative of #(sinx)*(cosx)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Shwetank Mauria Feb 5, 2018 #intsinxcosxdx=-1/2cos2x# Explanation: Antiderivative of #sinxcosx# means #intsinxcosxdx# = #intsin2xdx# and let #u=2x# then #du=(dx)/2# and hence #intsin2xdx=1/2intsinudu# = #-1/2cosu=-1/2cos2x# 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 10268 views around the world You can reuse this answer Creative Commons License