calculus - Compute $int_0^{pi/2}frac{cos{x}}{2-sin{2x}}dx$

How can I evaluate the following integral?

$$I=\int_0^{\pi/2}\frac{\cos{x}}{2-\sin{2x}}dx$$

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I tried it with Wolfram Alpha, it gave me a numerical solution: $0.785398$.
Although I immediately know that it is equal to $\pi /4$, I fail to obtain the answer with pen and paper.
I tried to use substitution $u=\tan{x}$, but I failed because the upper limit of the integral is $\pi/2$ and $\tan{\pi/2}$ is undefined.
So how are we going to evaluate this integral? Thanks.

Answer

Hint:

Knowing that $\sin2x=2\sin x\cos x$ and $\sin^2x+\cos^2x=1$. The integral can be expressed as

\begin{equation}
I=\int_0^{\pi/2}\frac{\cos x}{1+(\sin x-\cos x)^2}\ dx
\end{equation}

then use substitution $x\mapsto\frac{\pi}{2}-x$, we have

\begin{equation}
I=\int_0^{\pi/2}\frac{\sin x}{1+(\sin x-\cos x)^2}\ dx
\end{equation}

Add the two $I$'s and let $u=\sin x-\cos x$.