What is the integral of cos^2(6x) dx? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Ratnaker Mehta Jun 20, 2016 :. intcos^2(6x)dx=(1/2)(x+(sin12x)/12)+C. Explanation: Recall that cos^2theta=(1+cos2theta)/2. :. cos^2(6x)=(1+cos12x)/2. :. intcos^2(6x)dx=int(1+cos12x)/2dx=(1/2){x+(sin12x)/12}+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 28034 views around the world You can reuse this answer Creative Commons License