How to integrate $\frac {\cos (7x)-\cos (8x)}{1+2\cos (5x)} $ ?

integration

The rule is to multiply above and below by the sin(5x) (here).

$${\sin(5x) (\cos (7x)-\cos (8x))\over \sin(5x)+\sin(10x)}$$ $$={\sin(5x)2\sin(15x/2)\sin(x/2)\over 2\sin(15x/2)\cos(5x/2)}$$ $$=2\sin(5x/2)\sin(x/2)$$

Now its easy integration right?


HINT:

$$\dfrac{\cos y-\cos(6A+y)}{1+2\cos2A}=\dfrac{2\sin(3A+y)\sin3A}{1+2(1-2\sin^2A)}$$ $$=\dfrac{2\sin(3A+y)\sin A(3-4\sin^2A)}{3-4\sin^2A}=\cos(2A+y)-\cos(4A+y)$$

Here $2A=5x,6A+y=8x\implies y=-7x$

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