# 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: $$\frac{1}{20} \ln \left({3}\right)$$ $$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx =$$ $$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 32 \sin\left(x\right) \cos\left(x\right)} \, dx$$

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 $$\int du / u = \ln(u)+k$$.

$$I=\int_{0}^{\pi/4}\frac{\sin x +\cos x}{9+16 \sin 2x} dx.$$ 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 $$I=\frac{1}{16}\int_{-1}^{0} \frac{dt}{25/16-t^2}=\left .\frac{1}{16} \frac{4}{10}\log\frac{5/4+t}{5/4-t} \right|_{-1}^{0}= \frac{1}{40} \log 9=\frac{1}{20}\log3.$$

HINT

Apply a change of variables to get the interval $$[0,2\pi]$$ for integration and then

put $$z=e^{it}$$

Thus $$\cos{at}=\frac{z^a+\frac{1}{z^a}}{2}$$ and $$\sin{at}=\frac{z^a-\frac{1}{z^a}}{2i}$$

$$\frac{dz}{dt}=iz$$

Then apply the Residue Theorem to the unit circle.

You have a rational function of trigonometric functions. Apply the Weierstrassian substitution. That is, let $$\sin x=\frac{2t}{1-t^2}$$ and $$\cos x=\frac{1-t^2}{1+t^2},$$ where $$t=\tan(x/2).$$