Help me please? with indefinite integral ln(cosx)dx/cos^2x

1 Answer
Feb 18, 2018

I=ln(cos(x))tan(x)+tan(x)-x+C

Explanation:

We want to solve

I=intln(cos(x))/cos^2(x)dx=intln(cos(x))sec^2(x)dx

Use integration by parts

intudv=uv-intvdu

Let u=ln(cos(x)) and dv=sec^2(x)dx

Then du=-tan(x)dx and v=tan(x)

I=ln(cos(x))tan(x)-inttan(x)(-tan(x))dx

=ln(cos(x))tan(x)+inttan^2(x)dx

Then solve the integral

I_1=inttan^2(x)dx

Use the identity tan^2(x)=sec^2(x)-1

I_1=intsec^2(x)dx-int1dx=tan(x)-x+C

Therefore

I=ln(cos(x))tan(x)+tan(x)-x+C