What is the integral of #cos^2 6x#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer bp Apr 11, 2016 #1/2( x+1/12 sin 12x)# +C Explanation: #cos^2 6x = 1/2 (1+cos 12x)# {identity #cos2x= 2cos^2x -1#} #int cos^2 6x= int 1/2 (1+cos 12x) dx # =#1/2( x+1/12 sin 12x)# +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 6144 views around the world You can reuse this answer Creative Commons License