Question

${\int }_0^{2\pi }si{n}^4$ x dx is equal to

• A. $2{\int }_0^{\pi }si{n}^4$ x dc
• B. $8{\int }_0^{\frac{\pi }{4}}si{n}^4$ x dx
• C. $4{\int }_0^{\frac{\pi }{2}}co{s}^4$ x dx
• D. $3{\int }_0^{\frac{2\pi }{3}}si{n}^4$ x dx

Answer 1

Answer: (A), (C)

$4{\int }_0^{\frac{\pi }{2}}co{s}^4$ x dx
$2{\int }_0^{\pi }si{n}^4$ x dc

The period of $sin\left(x\right)=2\pi$ while $f\left(x\right)=si{n}^2\left(x\right)$ has period of $\pi$.
Hence
${\int }_0^{2\pi }si{n}^4\left(x\right).dx$
$={\int }_0^{2\pi }si{n}^4(2\pi -x).dx={\int }_0^{2\pi }sin\left(x\right).dx$
$=2{\int }_0^{\pi }si{n}^4\left(x\right).dx$
$=4{\int }_0^{\frac{\pi }{2}}si{n}^4\left(x\right).dx$
$=4{\int }_0^{\frac{\pi }{2}}si{n}^4(\frac{\pi }{2}-x).dx$
$=4{\int }_0^{\frac{\pi }{2}}co{s}^4\left(x\right).dx$