What is #int 1+cos2x dx #? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Leland Adriano Alejandro Jan 12, 2016 #x+1/2sin 2x + C# Explanation: if the given is #int (1+ cos 2x#) #dx# then #int (1+ cos 2x#) #dx#=#int dx# + #1/2int cos 2x # #2*dx + C# then #x + 1/2 sin 2x + 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 13502 views around the world You can reuse this answer Creative Commons License