$\int \sin ^3 x \cos ^2 x d x=$

• Correct answer
  A

  $\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}-\frac{2 \cos x}{15}+c$
• B

  $-\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}+\frac{2 \cos x}{15}+c$
• C

  $\frac{\sin ^4 x \cos x}{5}-\frac{\sin ^2 x \cos x}{15}+\frac{2 x}{15}+c$
• D

  $\frac{\sin ^4 x \cos x}{5}+\frac{\sin ^2 x \cos x}{3}-\frac{2 x}{15}+c$

Hence, the correct answer is option: 'A'.