Find the integral $\int_{0}^{\frac{\pi}{4}}\frac{\sin{x}\cos{x}}{\sin{x}+\cos{x}}dx$ find the integeral 
$$\int_{0}^{\frac{\pi}{4}}\dfrac{\sin{x}\cos{x}}{\sin{x}+\cos{x}}dx$$
I know 
$$\dfrac{2\sin{x}\cos{x}}{\sin{x}+\cos{x}}=\dfrac{(\sin{x}+\cos{x})^2-1}{\sin{x}+\cos{x}}=\sin{x}+\cos{x}-\dfrac{1}{\sin{x}+\cos{x}}$$
and
$$\int(\sin{x}+\cos{x})dx=\sin{x}-\cos{x}+C$$
and
$$\int\dfrac{1}{\sin{x}+\cos{x}}dx=\int\dfrac{1}{\sqrt{2}}\csc{(x+\dfrac{\pi}{4})}dx=\dfrac{1}{\sqrt{2}}\cdot(-\ln{(\cot{x}-\csc{x})})+C$$
I think this intgral have other methods.because this is simple form,
so My Question:
can you have other methods?

Here is a very slightly (and I mean very slightly) different approach compared to those already given.
Recognising
$$\sin x + \cos x = \sqrt{2} \sin \left (x + \frac{\pi}{4} \right ).$$
Now
\begin{align*}
(\sin x + \cos x)^2 &= \sin^2 x + 2 \sin x \cos x + cos^2 x\\
\Rightarrow 2 \sin^2 \left (x + \frac{\pi}{4} \right ) &= 1 + 2 \sin x \cos x\\
\Rightarrow \sin x \cos x &= \sin^2 \left (x + \frac{\pi}{4} \right ) - \frac{1}{2}.
\end{align*}
So our integral becomes
\begin{align*}
\int^{\frac{\pi}{4}}_0 \frac{\sin x \cos x}{\sin x + \cos x} \,dx &= \int^{\frac{\pi}{4}}_0 \frac{\sin^2 \left (x + \frac{\pi}{4} \right ) - \frac{1}{2}}{\sqrt{2} \sin \left (x + \frac{\pi}{4} \right )} \, dx\\
&= \frac{1}{\sqrt{2}} \int^{\frac{\pi}{4}}_0 \sin \left (x + \frac{\pi}{4} \right ) \, dx - \frac{1}{2 \sqrt{2}} \int^{\frac{\pi}{4}}_0 \text{cosec} \left (x + \frac{\pi}{4} \right ) \, dx\\
&= \frac{1}{\sqrt{2}} \left [- \cos \left (x + \frac{\pi}{4} \right ) \right ]^{\pi/4}_0 + \frac{1}{2\sqrt{2}} \ln \left (\text{cosec} \left (x + \frac{\pi}{4} \right ) + \cot \left (x + \frac{\pi}{4} \right ) \right ) \Big{|}^{\pi/4}_0\\
&= \frac{1}{2} - \frac{1}{2\sqrt{2}} \ln (1 + \sqrt{2}).
\end{align*}
