Evaluate the following integral: $\int_{0}^{\pi/4} \frac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx$ This is a question for the 2006 MIT Integration Bee that went unanswered by the contestants. I am not sure how to solve it either. I was only able to use the double angle formula to simplify the integral: $\sin\left({2x}\right) = 2 \sin\left(x\right) \cos\left(x\right)$
The final answer given was: $\frac{1}{20} \ln \left({3}\right)$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx = $$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 32 \sin\left(x\right) \cos\left(x\right)} \, dx $$

$$I=\int_{0}^{\pi/4}\frac{\sin x +\cos x}{9+16 \sin 2x} dx.$$
Use $\sin 2x=1-(\sin x- \cos x)^2$ and re-write 
$$I=\int_{0}^{\pi/4} \frac{\sin x +\cos x}{25-16(\sin x -\cos x)^2} dx,$$
Now use $\sin x -\cos x=t \implies (\cos x+ \sin x ) dx=dt$, then
$$I=\frac{1}{16}\int_{-1}^{0} \frac{dt}{25/16-t^2}=\left .\frac{1}{16} \frac{4}{10}\log\frac{5/4+t}{5/4-t} \right|_{-1}^{0}= \frac{1}{40} \log 9=\frac{1}{20}\log3.$$
