$ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \displaystyle &=\int_{\frac{\pi}{4}}^{0} \frac{\sin \left(\frac{\pi}{4}-y\right)+\cos \left(\frac{\pi}{4}-y\right)}{9+16 \sin 2\left[(\frac{\pi}{4}-y)\right]}(-d y) \\
\displaystyle &=\int_{0}^{\frac{\pi}{4}} \frac{\frac{1}{\sqrt{2}}(\cos y-\sin y)+\frac{1}{\sqrt{2}}(\cos y+\sin y)}{9+16 \cos 2 y} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\cos y}{9+16\left(1-2 \sin ^{2} y\right)} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{1}{\sqrt 2} } \frac{d z}{25-32 z^{2}} \text { , where } z=\sin y\\
\displaystyle &=\frac{\sqrt{2}}{10} \int_{0}^{\frac{1}{\sqrt 2} }\left(\frac{1}{5-4 \sqrt{2}z}+\frac{1}{5+4 \sqrt{2} z}\right) d z \\
\displaystyle &=\frac{\sqrt{2}}{10(4 \sqrt{2})}\left[\ln \left|\frac{5+4 \sqrt{2} z}{5-4 \sqrt{2} z}\right|\right]_{0}^{\frac{1}{\sqrt 2} } \\
\displaystyle &=\frac{1}{40}\ln 9 \end{aligned}$$
Best Answer
$$ \begin{aligned} \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x =& \int_{0}^{\frac{\pi}{4}} \frac{d(\sin x-\cos x)}{5^{2}-4^{2}(\sin x-\cos x)^{2}} \\ =& \frac{1}{40}\left[\ln \left| \frac{4 (\sin x-\cos x)+5}{4(\sin x-\cos x)-5} \right|\right]_{0}^{\frac{\pi}{4}}\\ =& \frac{\ln 9}{40} \end{aligned} $$