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**Prove It** $\displaystyle \displaystyle \int{\frac{1}{\sin{x}\cos{x}}\,dx} = \int{\frac{\cos{x}}{\sin{x}\cos^2{x}}\,dx}$

$\displaystyle \displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin^2{x})}\,dx}$

$\displaystyle \displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin{x})(1 + \sin{x})}\,dx}$.

Now make the substitution $\displaystyle \displaystyle u = \sin{x}$ so that $\displaystyle \displaystyle \frac{du}{dx} = \cos{x}$ and the integral becomes

$\displaystyle \displaystyle \int{\frac{1}{u(1 - u)(1 + u)}\,\frac{du}{dx}\,dx}$

$\displaystyle \displaystyle = \int{\frac{1}{u(1 - u)(1 + u)}\,du}$

$\displaystyle \displaystyle = \int{\frac{1}{u} - \frac{1}{2(u - 1)} - \frac{1}{2(u + 1)}\,du}$

$\displaystyle \displaystyle = \ln{|u|} - \frac{1}{2}\ln{|u-1|} - \frac{1}{2}\ln{|u + 1|} + C$

$\displaystyle \displaystyle = \ln{|\sin{x}|} - \frac{1}{2}\ln{|\sin{x} - 1|} - \frac{1}{2}\ln{|\sin{x} + 1|} + C$.