Question

${\int }_0^{\pi /4}{\sin }^{4}x.{\cos }^{4}xdx=$

• A. $\frac{3\pi }{512}$
• B. $\frac{\pi }{512}$
• C. $\frac{-{\pi }^2}{512}$
• D. $\frac{7\pi }{512}$

Answer 1

Answer: (A)

$\frac{3\pi }{512}$

${\int }_0^{\frac{\pi }{4}}si{n}^{4}xco{s}^{4}xdx$
$I_{m,n}=\frac{(m-1)}{(m+n)}\cdot \frac{(m-3)}{(m+n-2)}\cdot \frac{(m-5)}{(m+n-4)}.....I_{0,4}$
$=(\frac{3}{8}\times \frac{1}{6})I_{0,4}$
$I_{0,4}={\int }_0^{\frac{\pi }{2}}co{s}^{4}xdx=\frac{3}{4}\times \frac{1}{4}\times \frac{\pi }{2}$
$=\frac{3\pi }{16}$
$I_{4,4}=\frac{3\pi }{16}(\frac{3}{8}\times \frac{1}{6})=\frac{3\pi }{512}$