How do you find the integral of #int 1/(1 + csc(x))#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Dharma R. Sep 18, 2015 #x+secx-tanx+c# Explanation: # int1/(1+cscx)dx=intsinx/(1+sinx)dx=int(sinx-sin^2x)/(cos^2x)dx# =#inttanxsecxdx-inttan^2xdx# =#inttanxsecxdx-int (sec^2x-1)dx# =#secx-(tanx-x)+c# =#x+secx-tanx+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 5293 views around the world You can reuse this answer Creative Commons License