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

All I could do is apply difference of cosines formula in numerator.After that I'm stuck.Can somebody please help me?

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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?

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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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Let $$I = \int\frac{\cos 7x-\cos 8x}{1+2\cos 5x}dx = \int\frac{(\cos 7x+\cos 3x)-(\cos 8x+\cos 2x)-\cos 3x+\cos 2x}{1+2\cos 5x}dx$$

$$\bullet \displaystyle \cos C+\cos D = 2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$$

So $$I = \int\frac{2\cos 5x\cos 2x+\cos 2x-2\cos 5x\cos3x-\cos 3x}{1+2\cos 5x}dx$$

$$I = \int\frac{(2\cos 5x+1)(\cos 2x-\cos 3x)}{1+2\cos 5x}dx = \int(\cos 2x-\cos 3x )dx$$

So $$I = \frac{\sin 2x}{2}-\frac{\sin 3x}{3}+\mathcal{C}$$

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