Is there another way to solve this integral?

Substitute $x=\pi/2-2t$ so the integral becomes $$ -2\int_{\pi/4}^{-\pi/4}\frac{1-\cos 2t}{1+\cos 2t}\,dt= 2\int_{-\pi/4}^{\pi/4}\frac{1-\cos^2t}{\cos^2t}\,dt =2\Bigl[\tan t - t\Bigr]_{-\pi/4}^{\pi/4} $$

First use partial fractions to get rid of the sine in the numerator: $$ \int_0^{\pi} \frac{1-\sin{x}-1+1}{1+\sin{x}} \, dx = \int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, dx = -\pi + \int_0^{\pi} \frac{2 \, dx}{1+\sin{x}}. $$ We have the identity $$ 1+\sin{x} = 2\sin^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} $$ (from $\cos{2\theta}=1-2\sin^2{\theta}$ and $\sin{\theta}=-\cos{(\theta+\pi/2)}$), so the remaining integral is $$ \int_0^{\pi} \csc^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \, dx = \left[ -2 \cot{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \right]_0^{\pi} = -2(-1-1) = 4 $$

Multiply numerator and denominator by $1 - \sin x$. So that $\displaystyle\int_0^\pi \dfrac{1-\sin x}{1+\sin x} \cdot \dfrac{1-\sin x}{1-\sin x}dx $

$$ = \displaystyle\int_0^\pi \dfrac{1-2\sin x+ \sin^2x}{\cos^2x}dx = \int_0^\pi \sec^2x - 2\dfrac{\sin x}{\cos^2x} + \tan^2x dx$$

We know that $\tan^2x + 1 = \sec^2x$. So the integral would look like: $$= \int_0^\pi 2\sec^2x -1 - 2\dfrac{\sin x}{\cos^2x}dx$$

We know the integral of $\sec^2x$ is just $\tan x$, and the last integrand can be solved by letting $u=\cos x$.