$\int \frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x} d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} I & =\int \frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x} d x \ & =\int \frac{\cos 7 x-\cos 8 x}{1+2-4 \sin ^2\left(\frac{5 x}{2}\right)} d x \ & =\int \frac{2 \sin \frac{15 x}{2} \sin \frac{x}{2}}{3-4 \sin ^2\left(\frac{5 x}{2}\right)} d x \ & =\int \frac{2 \sin \frac{15 x}{2} \sin \frac{x}{2} \sin \frac{5 x}{2}}{3 \sin \frac{5 x}{2}-4 \sin ^3 \frac{5 x}{2}} d x \ & \left.=\int 2 \sin \frac{x}{2} \sin \frac{5 x}{2} d x=\int \cos 2 x-\cos 3 x\right) d x \ & =\frac{1}{2} \sin 2 x-\frac{1}{3} \sin 3 x+c \end{aligned}$

\(\int \frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x} d x=\)

Solution