Help me please? with indefinite integral ln(cosx)dx/cos^2x
1 Answer
Feb 18, 2018
Explanation:
We want to solve
I=intln(cos(x))/cos^2(x)dx=intln(cos(x))sec^2(x)dx
intudv=uv-intvdu
Let
Then
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
I_1=intsec^2(x)dx-int1dx=tan(x)-x+C
Therefore
I=ln(cos(x))tan(x)+tan(x)-x+C