Question

$ \int{({{\sin }^{6}}x+{{\cos }^{6}}x+3{{\sin }^{2}}x \,{{\cos }^{2}}x)}dx $ is equal to

• A. $ x+c $
• B. $ \frac{3}{2}\sin 2x+c $
• C. $ -\frac{3}{2}\cos 2x+c $
• D. $ \frac{1}{3}\sin 3x-\cos 3x+c $
• E. $ \frac{1}{3}\sin 3x+\cos 3x+c $

Answer 1

Answer: (A)

Let $ I=\int{({{\sin }^{6}}x+{{\cos }^{6}}x+3{{\sin }^{2}}x{{\cos }^{2}}x)}dx $
$=\int{\{{{({{\sin }^{2}}x)}^{3}}+{{({{\cos }^{2}}x)}^{3}}} $ $ +3{{\sin }^{2}}x{{\cos }^{2}}x\}dx\} $
$=\int{\left[ \begin{align} & ({{\sin }^{2}}x+{{\cos }^{2}}x)({{\sin }^{4}}x+{{\cos }^{4}}x \\ & -{{\sin }^{2}}x{{\cos }^{2}}x)+3{{\sin }^{2}}x{{\cos }^{2}}x \\ \end{align} \right]}dx $
$=\int{\left[ \begin{align} & {{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-3{{\sin }^{2}}x{{\cos }^{2}}x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3{{\sin }^{2}}x{{\cos }^{2}}x \\ \end{align} \right]}dx $
$=\int{1\,dx}$
$ = x+c $