solving an integral calculus equation with Natural Logarithm cube

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I'm trying to understand this equation here, Is it possible to solve an equation with natural logarithm without plugin a value into it? on this equation, it's just $\ln^3$ without any value, or is this a type error and should be written like this $\ln(3)$? I'm sorry, if this question sounds dumb, I'm trying to advance learn integral calculus. An answer would be so much appreciated. Thanks

$$ \int \frac{2}{\cos(x)} \ln^3 \left[\frac{1-\sin(x)}{1+\sin(x)}\right] \ \text{d}x$$

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It is not true that $\ln^3$ appears “without any value” here. The author is applying $\ln^3$ to $\frac{1-\sin x}{1+\sin x}$. Another way of writing that integral is$$\int\frac2{\cos x}\ln^3\left(\frac{1-\sin x}{1+\sin x}\right)\,\mathrm dx.$$