Is there a nicer way to compute this integral?? What's the trick to this?
Here's my attempt:
$$\int_{0}^{\pi}\sin(x)\sqrt{\frac{1+\cos(9x)}{1+\cos(x)}}dx=\int_{0}^{\pi} \tan(x)\cos(\frac{9x}{2})dx=\frac{1}{2}\int_{0}^{\pi}\frac{\sin(\frac{11x}{2})-\sin(\frac{7x}{2})}{\cos(x)}dx$$
As you can see, this is not a pretty integral.
Best Answer
Using double angle identities the integrand simplifies to
$$\int_0^\pi 2\sin\frac{x}{2}\left|\cos\frac{9x}{2}\right|dx = \int_0^\pi |\sin 5x - \sin 4x |dx$$
The absolute value in the denominator can be ignored when simplifying since it is always nonnegative on the specified domain. However this integral's answer is the messy
$$\frac{1}{20}\left(17+18\cos\frac{\pi}{18}+18\cos\frac{\pi}{9}+18\cos\frac{2\pi}{9}\right)$$
after taking into account the zero crossings of the integrand. The answer without the absolute value would have been $\frac{1}{5}$, and maybe that was a mistake on the problem writer's part.