Help with Definite integral question Anyone please help with this question:
(a) Show that: 
\begin{align}
 \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx 
\end{align}
(b) Hence show that:
\begin{align} 
\int_{0}^{\frac{\pi}{4}} \frac{1-\sin(2x)}{1+\sin(2x)} dx = \int_{0}^{\frac{\pi}{4}} \tan^2{x} dx 
\end{align}
And evaluate the integral.
I'd done part (a) and for part (b) I tried:
\begin{align}
\int_{0}^{\frac{\pi}{4}} \frac{1-\sin(2(\frac{\pi}{4} - x))}{1+\sin(2(\frac{\pi}{4} - x))}
\end{align}
\begin{align}
\int_{0}^{\frac{\pi}{4}} \frac{1-\sin\left(\frac{\pi}{2} - 2x\right)}{1+\sin \left(\frac{\pi}{2} - 2x\right)}
\end{align}
Which give: 
\begin{align}
\int_{0}^{\frac{\pi}{4}} \frac{1-\cos(2x)}{1+\cos(2x)}
\end{align}
But I was lost after that. I'm not sure if I'm on the right track...
Thanks for your time.

This is a H/W question from P. 173 of Arnold and Arnold. 4 Unit Mathematics. Melbourne 1993. 

Using the double angle identities to express $\cos 2x$ in terms of $\cos x$ and $\sin x$, we get: 
$$\frac{1-\cos 2x}{1+\cos 2x} = \frac{1- (1 - 2 \sin^2 x)}{1 + (2 \cos^2 x - 1)} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$$
