Final answer to the problem
$y=\frac{1}{2}\ln\left(\frac{1+\sin\left(x\right)}{1-\sin\left(x\right)}\right)$
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Apply the formula: $\ln\left(x^a\right)$$=a\ln\left(x\right)$, where $a=\frac{1}{2}$ and $x=\frac{1+\sin\left(x\right)}{1-\sin\left(x\right)}$
$y=\frac{1}{2}\ln\left(\frac{1+\sin\left(x\right)}{1-\sin\left(x\right)}\right)$
Final answer to the problem
$y=\frac{1}{2}\ln\left(\frac{1+\sin\left(x\right)}{1-\sin\left(x\right)}\right)$
