If $\mathrm{I}_1=\int \sin ^6 x d x$ and $\mathrm{I}_2=\int \cos ^6 x d x$ then $\mathrm{I}_1+\mathrm{I}_2=$

• A

  $\frac{5 x}{8}+\frac{3 \cos 4 x}{32}+c$
• B

  $\frac{1}{32}(20 x-3 \sin 4 x)+c$
• Correct answer
  C

  $\frac{1}{32}(20 x+3 \sin 4 x)+c$
• D

  $\frac{5 x}{4}+\frac{3 \sin 4 x}{16}+c$

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