Q.Integrate: $\int \frac{\cos 7x-\cos 8x}{1+2\cos 5x}\,dx$

Numerator: $\cos7x-\cos8x=2\sin\left(\frac{15x}{2}\right)\sin\left(\frac{x}{2}\right)$ (sum-to-product, $\cos A-\cos B=2\sin\frac{A+B}2\sin\frac{B-A}2$). Denominator: using the identity $\sin\frac{3\theta}{2}=\sin\frac\theta2\,(1+2\cos\theta)$ with $\theta=5x$, we get $1+2\cos5x=\dfrac{\sin(15x/2)}{\sin(5x/2)}$. So the whole fraction is $\dfrac{2\sin(15x/2)\sin(x/2)}{\sin(15x/2)/\sin(5x/2)}=2\sin(x/2)\sin(5x/2)$ — the common factor $\sin(15x/2)$ cancels. Applying $2\sin A\sin B=\cos(A-B)-\cos(A+B)$ with $A=5x/2,B=x/2$: $2\sin(5x/2)\sin(x/2)=\cos2x …

Sum-to-product on the numerator, the identity $1+2\cos\theta=\sin(3\theta/2)/\sin(\theta/2) …

Not knowing the $1+2\cos\theta$ identity and instead trying to expand $\cos5x$ in terms of $\cos x$ via multiple-angle formulas, which becomes intractable …