${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{2}xco{s}^{2}xdx=$

Hint: you may $($or may not$)$ need to use any of the following reduction formulae:

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{s}^{n}xdx=\frac{co{s}^{n-1}xsinx}{n}+\frac{n-1}{n}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{s}^{n-2}xdx$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{n}xdx=-\frac{si{n}^{n-1}xcosx}{n}+\frac{n-1}{n}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{n-2}xdx$

More than one answer may be correct.

$\frac{1}{8}x+\frac{1}{16}sin2x-\frac{1}{4}sinxco{s}^{3}x+C$

$\frac{1}{8}x-\frac{1}{32}sin4x+C$

$\frac{1}{8}x+\frac{1}{32}cos4x+C$

$\frac{1}{8}x-\frac{1}{16}cos2x+\frac{1}{4}si{n}^{3}xcosx+C$

$\frac{1}{8}x-\frac{1}{16}sin2x+\frac{1}{4}si{n}^{3}xcosx+C$

$\frac{1}{8}x+\frac{1}{16}cos2x-\frac{1}{4}sinxco{s}^{3}x+C$

Attention: this answer may contain $2$ or more answers. For example, answer option $2$ is correct. However, coursera will count it as incorrect, since there are still possible options