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 1491 views around the world You can reuse this answer Creative Commons License