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