What is the antiderivative of #cos^2 x#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer VNVDVI May 13, 2018 #x/2+1/4sin2x+C# Explanation: Recall the cosine power reduction formula, #cos^2x=1/2(1+cos2x)# Then, #intcos^2xdx=1/2int(1+cos2xdx)=1/2(intdx+intcos2xdx)# #=1/2(x+1/2sin2x+C)=x/2+1/4sin2x+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 15812 views around the world You can reuse this answer Creative Commons License