## Calculus: Early Transcendentals 8th Edition

$$\int\sin^2x\cos^3 xdx=\frac{\sin^3x}{3}-\frac{\sin^5x}{5}+C$$
$$A=\int\sin^2x\cos^3 xdx$$ Since the power of $\cos$ is odd, the strategy is to save 1 $\cos$ factor and change the rest in terms of $\sin$ according to the formula $\cos^2 x=1-\sin^2x$ Therefore, we have $$A=\int\sin^2x\cos^2x\cos xdx$$ $$A=\int\sin^2x(1-\sin^2 x)\cos xdx$$ Now we substitute $u=sin x$ That makes $du=\cos xdx$ So, $$A=\int u^2(1-u^2)du$$ $$A=\int (u^2-u^4)du$$$$A=\frac{u^3}{3}-\frac{u^5}{5}+C$$ $$A=\frac{\sin^3x}{3}-\frac{\sin^5x}{5}+C$$