Question

If ${\int }_0^{\pi /2}{\sin }^{6}xdx=\frac{5\pi }{32},$ then the value of ${\int }_{-\pi }^{\pi }({\sin }^{6}x+{\cos }^{6}x)dx$ is

• A. $5\pi /8$
• B. $5\pi /16$
• C. $5\pi /2$
• D. $5\pi /4$
• E. $5\pi /32$

Answer 1

Answer: (D)

Given that, ${\int }_0^{\pi /2}{\sin }^{6}xdx=\frac{5\pi }{32}$
Let $I={\int }_{-\pi }^{\pi }({\sin }^{6}x+{\cos }^{6}x)dx$
$=2{\int }_0^{\pi }({\sin }^{6}x+{\cos }^{6}x)dx$
$=4{\int }_0^{\pi /2}({\sin }^{6}x+{\cos }^{6}x)dx$
$=4{\int }_0^{\pi /2}{\sin }^{6}xdx+4{\int }_0^{\pi /2}{\cos }^{6}x\left(\frac{\pi }{2}-x\right)dx$
$=8{\int }_0^{\pi /2}{\sin }^{6}xdx+8\times \frac{\mathrm{5.3.1}}{\mathrm{6.4.2}}\times \frac{\pi }{2}$
$=\frac{5\pi }{4}$