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. $\frac{1}{3}\sin 3x-\cos 3x+c$

Answer 1

Answer: (A)

$x+c$

Answer 2

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