## SUBJECT 3.

Calculate:
a) $\int \frac{\sin 7 x}{\sin x} d x, \mathrm{x} \in\left(0, \frac{\pi}{2}\right)$
b) $\int \frac{\cos 7 x}{\cos x} d x, \mathrm{x} \in\left(0, \frac{\pi}{2}\right)$
## Subject 3.

$\mathrm{I}+\mathrm{J}=\int\left(\frac{\sin 7 x}{\sin x}+\frac{\cos 7 x}{\cos x}\right) d x=\int \frac{\sin 8 x}{\sin x \cos x} d x=2 \int \frac{\sin 8 x}{\sin 2 x} d x=2 \int \frac{2 \sin 4 x \cos 4 x}{\sin 2 x} d x=4 \int 2 \cos 2 x \cos 4 x d x=$ $=8 \int \frac{\cos 6 x+\cos 2 x}{2} d x=4 \frac{\sin 6 x}{6}+4 \frac{\sin 2 x}{2}=\frac{2}{3} \sin 6 x+2 \sin 2 x+C$

$\mathrm{I}-\mathrm{J}=\int\left(\frac{\sin 7 x}{\sin x}-\frac{\cos 7 x}{\cos x}\right) d x=\int \frac{\sin 6 x}{\sin x \cos x} d x=2 \int \frac{\sin 6 x}{\sin 2 x} d x=2 \int \frac{3 \sin 2 x-4 \sin ^{3} 2 x}{\sin 2 x} d x=2 \int 3-4 \sin ^{2} 2 x d x=$ $=6 x-8 \int \sin ^{2} 2 x d x=6 x-8 \int \frac{1-\cos 4 x}{2} d x=2 x+\sin 4 x+C$

. $3 \mathbf{p}$

From (1) and (2) it follows that $\mathrm{I}=\frac{1}{3} \sin 6 x+\sin 2 x+x+\frac{1}{2} \sin 4 x+C$ and $\mathrm{J}=\frac{1}{3} \sin 6 x+\sin 2 x-x-\frac{1}{2} \sin 4 x+C$

$1 p$