How do you find the integral of int (sin x)/(cos^2x + 1)dx? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Sasha P. Oct 20, 2015 I = -arctan (cosx) + C Explanation: I = int sinx/(cos^2x+1)dx = int (sinxdx)/(cos^2x+1) = -int (d(cosx))/(cos^2x+1) I = -arctan (cosx) + 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 1652 views around the world You can reuse this answer Creative Commons License