Cos, it's deceiving

Using the identity

$\cos C-\cos D=2\sin\dfrac{C+D}{2}\sin \dfrac{D-C}{2}$

The numerator simplifies to $-2\sin(15x/2)\sin(x/2)$ .

Using the identity

$\dfrac{\sin 3\theta}{\sin\theta}=1+2\cos 2\theta$

and putting $\theta=5x/2$ the denominator simplifies to $\dfrac{\sin (15x/2)}{\sin(5x/2)}$ .

Thus the integral becomes

$\displaystyle\int_{0}^{\pi/2} \cos 3x-\cos 2x\text{ }\text{d}x$

(after applying the first identity a second time) which evaluates to $\boxed{\dfrac{-1}{3}}$

Well, I don't know a simpler solution than this.It would be nice if I could get one.

$\displaystyle \begin{array}{c}\\ I && = \int_0^{\frac{\pi }{2}} \frac{\cos (8x) - \cos (7x)}{1 + 2 \cos (5x) } \text{ d}x \\ && = \int_0^{\frac{\pi }{2}} \frac{\left( \cos (8x) + \cos (2x)\right) - \left( \cos (7x) + \cos(3x) \right) + \left( \cos(3x) - \cos(2x) \right)}{1 + 2 \cos (5x) } \text{ d}x \\ && = \int_0^{\frac{\pi }{2}} \frac{2 \cos (5x) \cos (3x) - 2 \cos (5x) \cos (2x) + \left( \cos(3x) - \cos(2x) \right) }{1 + 2 \cos (5x) } \text{ d}x \\ && = \int_0^{\frac{\pi }{2}} \frac{\left( \cos(3x) - \cos(2x) \right) \left( 1 + 2 \cos(5x) \right)}{ 1 + 2 \cos (5x) } \text{ d}x \\ && = \int_0^{\frac{\pi }{2}} \cos (3x) - \cos (2x) \text{ d}x \\ && = \left[ \frac{ \sin (3x) }{3} - \frac{ \sin (2x) }{2} \right]_0^{\frac{\pi }{2}} \\ && = \frac{-1}{3} \\ \end{array}$

$\displaystyle \therefore 15I = \boxed{ -5 }$