Show that

Let $I=\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$

Also, let $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$

When $x=0, t=-1$ and when $x=\frac{\pi}{4}, t=0$

$\Rightarrow(\sin x-\cos x)^{2}=t^{2}$

$\Rightarrow \sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=t^{2}$

$\Rightarrow 1-\sin 2 x=t^{2}$

$\Rightarrow \sin 2 x=1-t^{2}$

$\therefore I=\int_{-1}^{0} \frac{d t}{9+16\left(1-t^{2}\right)}$

$=\int_{-1}^{0} \frac{d t}{9+16-16 t^{2}}$

$=\int_{-1}^{0} \frac{d t}{25-16 t^{2}}=\int_{-1}^{0} \frac{d t}{(5)^{2}-(4 t)^{2}}$

$=\frac{1}{4}\left[\frac{1}{2(5)} \log \left|\frac{5+4 t}{5-4 t}\right|\right]_{-1}^{0}$

$=\frac{1}{40}\left[\log (1)-\log \left|\frac{1}{9}\right|\right]$

$=\frac{1}{40} \log 9$