Answer
$$ x-\dfrac{7}{6} x^3+\dfrac{61}{120} x^5-\dfrac{1247}{5040} x^7+....$$
Work Step by Step
Recall the Taylor series for $\cos x=1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....$ and $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
$\cos^2 x \sin x =[1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....] \times [x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....]$
or, $\cos^2 x \sin x =x-\dfrac{7}{6} x^3+\dfrac{61}{120} x^5-\dfrac{1247}{5040} x^7+....$
