How do you find the antiderivative of #(1+cosx)^2#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Konstantinos Michailidis Jun 8, 2016 We have that #int (1+cosx)^2 dx=int (1+2*cosx+cos^2x)dx=int 1dx+2*int cosx+int cos^2x dx# First we notice that #cos^2 x=1/2(1+cos2x)# Hence #int (1+cosx)^2 dx=int (1+2*cosx+cos^2x)dx=int 1dx+2*int cosx+int cos^2xdx=x+2*sinx+x/2+1/4*sin2x+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 23255 views around the world You can reuse this answer Creative Commons License