Question #e2f56 Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Vishwaksen Reddy V. · NickTheTurtle Apr 6, 2017 #tan(x/2)+C# Explanation: #cos^2(x/2) = (1+cos(x))/2# So #1+cos(x) = 2cos^2(x/2)# and #1/(1+cos x) =1/2sec^2(x/2)# #int\ 1/2sec^2(x/2)\ dx = 1/2int\ sec^2(x/2)\ dx# Substituting #u=x/2# and #du=dx/2#, we get #1/2int\ 2sec^2(u)\ du=tan(u)+C=tan(x/2)+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 1328 views around the world You can reuse this answer Creative Commons License