## Calculus: Early Transcendentals 8th Edition

$$\int\sin8x\cos5xdx=-\frac{1}{26}\cos13x-\frac{1}{6}\cos3x+C$$
$$A=\int\sin8x\cos5xdx$$ For this problem, we need to utilize $$\sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$$ Therefore, $$A=\frac{1}{2}\int\sin(8x+5x)+\sin(8x-5x)dx$$ $$A=\frac{1}{2}\int(\sin13x+\sin3x)dx$$ $$A=\frac{1}{2}(\int\sin13xdx+\int\sin3xdx)$$ $$A=\frac{1}{2}[\frac{1}{13}\int\sin13xd(13x)+\frac{1}{3}\int\sin3xd(3x)]$$ $$A=\frac{1}{2}[-\frac{1}{13}\cos13 x-\frac{1}{3}\cos3x]+C$$ $$A=-\frac{1}{26}\cos13x-\frac{1}{6}\cos3x+C$$