## Calculus: Early Transcendentals 8th Edition

$$\int\frac{\sin 2x}{\sin x}dx=2\sin x+C$$
$$A=\int\frac{\sin 2x}{\sin x}dx$$ $$A=\int\frac{2\sin x\cos x}{\sin x}dx$$ $$A=\int2\cos xdx$$ From Table 1, $$\int cf(x)dx=c\int f(x)dx$$ Therefore, $$A=2\int\cos xdx$$ From Table 1, we also get $$\int(\cos x)dx=\sin x+C$$ Therefore, $$A=2\sin x+C$$