Integral: $$\int \ln(\cos^2 x) dx$$
I applied parts technique twice, first by expressing it as $-2\int1\cdot\ln(\sec x)$, which is
$$I = -2x\ln(\sec x)+2\int x\tan (x)dx$$
Now the second term can be integrated by parts to give:
$$I = x\ln(\cos^2 x) + 2\left(x\ln(\sec x)-\int\ln(\sec x)dx\right)$$
But second term of the bracket itself is $\dfrac{I}{2}$, so
$$I = x\ln(\cos^2 x) + 2x\ln(\sec x)+I$$
Now the I cancels out! I am not getting anywhere with this approach.
Best Answer
You integral is related with the inverse tangent integral and the dilogarithm function.
The substitutions $x=\arctan t$, $1+t^2=u$, $u=\frac{1}{v}$ bring your integral into $$ -\int \frac{\log(1+t^2)}{1+t^2}\,dt = -\int \frac{\log(u)\,du}{2u\sqrt{u-1}}=\int \frac{\log(v)\,dv}{\sqrt{v(1-v)}}$$ and the last integral is $$ \left.\frac{d}{d\alpha}\int v^{\alpha-\frac{1}{2}}(1-v)^{-1/2}\,dv\,\right|_{\alpha=0^+} $$ i.e. the derivative of an incomplete Beta function.