Question

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

Answer 1

`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`