The question is whether the two values of $I_2$ as obtained below are same? Integrate, $\int \frac{\sin^{2}{2\theta}}{2\cos{2\theta}}d\theta$
Let, $\sin2\theta=t$
then, ${2\cos2\theta} d\theta=dt$
$dt=\frac{1}{2}\frac{1}{\sqrt{1-t^2}}$
Integral =$\frac{1}{4}\int {\frac{t^2}{1-t^2}} dt$
Integral =$\frac{1}{4}\int {\frac{t^2}{1-t^2}} dt$
Integral =$-\frac{1}{4}\int {\frac{t^2-1+1}{t^2-1}} dt$
Integral =$-\frac{1}{4}\int ({1+\frac{1}{t^2-1}} )dt$
Integral =$-\frac{1}{4} (t+\int{\frac{1}{t^2-1}} dt)$
Integral =$-\frac{1}{4} (t+I_2)$
$I_2=\int \frac{1}{t^2-1}dt$
Now, $I_2=\frac{1}{2}\int (\frac{1}{t-1}-\frac{1}{t+1})dt$
Now, $I_2=\frac{1}{2}log(\frac{t-1}{t+1})+c_2dt$ ---(I)
we can also have the following,
$I_2=-\int \frac{1}{1-t^2}dt$
Now, $I_{2}=-\frac{1}{2}\int (\frac{1}{1+t}+\frac{1}{1-t})dt$
Now, $I_{2}=-\frac{1}{2}log(\frac{1+t}{1-t})+c_{22}$ ---(II)
The question is whether these two values of $I_2$ are same ? If, yes how do we show that both the values of $I_2$ as obtained in (I)&(II) are same ?

A few things to point out: 


*

*$\int\frac{1}{x}dx=\ln|x|+C$.

*$\ln\frac{a}{b}=-\ln\frac{b}{a}$.
