Question #e2f56 Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Vishwaksen Reddy V. ยท 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 1517 views around the world You can reuse this answer Creative Commons License