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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.

Max Payne
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    $$\log(\cos^2(x))=2\log(\cos(x))=2\log(2)+2\log(e^{ix}+e^{-ix})= 2\log(2)-2ix+2\log(e^{2ix}+1)=\ 2\log(2)-2ix+2\sum_{n\geq1}\frac{(-1)^ne^{ixn}}{n}$$

    also use the defintion of dilogarithm function

     – tired Jul 25 '16 at 12:50
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    There is a missing minus at the beginning of line three, this explains why they cancel out. You first integrated by parts then undid it. – entrelac Jul 25 '16 at 12:50
  • The integral of log(cos(x)) has no 'closed form' expression using standard functions. You will need to define it in terms of the series $x^2/n$ from n = 1 to infinity. – jg mr chapb Jul 25 '16 at 12:52
  • Related definite integral: http://math.stackexchange.com/q/690644/269624 – Yuriy S Jul 25 '16 at 14:09

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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.

Jack D'Aurizio
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