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