How do you integrate #cos2x#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer GiĆ³ Feb 17, 2015 You can set #2x=t# so that #x=t/2# and #dx=1/2dt# Your integral becomes: #intcos(t)/2dt=sin(t)/2+c# and going back to #x# #=sin(2x)/2+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 2784 views around the world You can reuse this answer Creative Commons License