<s>[INST] 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: $\frac{1}{20} \ln \left({3}\right)$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx = $$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 32 \sin\left(x\right) \cos\left(x\right)} \, dx $$
 [/INST] Note that \begin{align}
9+ 16\sin\left(2x\right)&=25+32\sin(x)\cos(x)-16\\
&=25-16\sin^2x-16\cos^2x+32\cos x\sin x\\
&=5^2-4^2(\sin x - \cos x)^2\\
&=(5-4\cos x+4\sin x)(5+4\cos x-4\sin x)
\end{align}
hence 
\begin{align}
\frac{\sin x+\cos x}{(5-4\cos x+4\sin x)(5+4\cos x-4\sin x)}&=-\frac{1}{40}\frac{-4\cos x-4\sin x}{(5+4\cos x-4\sin x)}\\
&+\frac{1}{40}\frac{4\cos x+4\sin x}{(5-4\cos x+4\sin x)}
\end{align}
... the rest is shall be manageable: for each fraction the numerator is the derivative of the denominator, hence you need to recall $\int du / u = \ln(u)+k$.
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