What is the integral of ln(cos(x))-ln(sin(x)) ?

The integral of ln(cos(x))-ln(sin(x)) is

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integral of ln(cos(x))-ln(sin(x))

\int ln (cos (x)) - ln (sin (x)) d x

\frac{1}{2} i Li_{2} (- e^{2 i x}) + \frac{i x ^{2}}{2} - x ln (1 + (- 1)^{\frac{2 x}{π}}) + x ln (cos (x)) - i \frac{1}{2} (x^{2} + Li_{2} (e^{2 i x}) - x ln (1 - (- 1)^{\frac{2 x}{π}}) + x ln (sin (x))) + C

Solution steps

\int ln (cos (x)) - ln (sin (x)) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int ln (cos (x)) d x - \int ln (sin (x)) d x

\int ln (cos (x)) d x = \frac{1}{2} i Li_{2} (- e^{2 i x}) + \frac{i x ^{2}}{2} - x ln (1 + (- 1)^{\frac{2 x}{π}}) + x ln (cos (x))

\int ln (sin (x)) d x = i \frac{1}{2} (x^{2} + Li_{2} (e^{2 i x}) - x ln (1 - (- 1)^{\frac{2 x}{π}}) + x ln (sin (x)))

= \frac{1}{2} i Li_{2} (- e^{2 i x}) + \frac{i x ^{2}}{2} - x ln (1 + (- 1)^{\frac{2 x}{π}}) + x ln (cos (x)) - i \frac{1}{2} (x^{2} + Li_{2} (e^{2 i x}) - x ln (1 - (- 1)^{\frac{2 x}{π}}) + x ln (sin (x)))

Add a constant to the solution

= \frac{1}{2} i Li_{2} (- e^{2 i x}) + \frac{i x ^{2}}{2} - x ln (1 + (- 1)^{\frac{2 x}{π}}) + x ln (cos (x)) - i \frac{1}{2} (x^{2} + Li_{2} (e^{2 i x}) - x ln (1 - (- 1)^{\frac{2 x}{π}}) + x ln (sin (x))) + C

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What is the integral of ln(cos(x))-ln(sin(x)) ?
The integral of ln(cos(x))-ln(sin(x)) is