What's the integral of int sinx * tanxdx? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Tom Nov 11, 2015 intsin(x)tan(x)dx = intsin^2(x)/cos(x)dx = int(sin^2(x)cos(x))/cos^2(x)dx let's t = sin(x) dt = cos(x) int(t^2)/(1-t^2)dt=int(t^2-1+1)/(1-t^2)dt = -int(1-t^2-1)/(1-t^2)dt =-(intdt-int1/(1-t^2)dt) =-[t]+[arctanh(t)]+C Substitute back for t=sin(x) -[sin(x)]+[arctanh(sin(x))]+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 1601 views around the world You can reuse this answer Creative Commons License