Question

Solve : $\frac{\pi }{128}<\underset{\pi /4}{\stackrel{\pi /2}{\int }}(\sin x{)}^{10}dx<\frac{\pi }{4}$

Answer 1

If $m$ and $M$ be global minima and global maxima of $f\left(x\right)$ in $[a,b]$ then

$m(b-a)\le {\int }_a^{b}f\left(x\right)dx\le M(b-a)$

$M=$ maximum of ${\sin }^{10}x={\sin }^{10}\frac{\pi }{2}=1$

$m=$ minimum of ${\sin }^{10}x={\sin }^{10}\frac{\pi }{4}={\left(\frac{1}{\sqrt{2}}\right)}^{10}=\frac{1}{32}$

Hence,

$\frac{1}{32}\left(\frac{\pi }{4}\right)\le {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\sin }^{10}xdx\le 1\left(\frac{\pi }{4}\right)$

$\left(\frac{\pi }{128}\right)\le {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\sin }^{10}xdx\le \left(\frac{\pi }{4}\right)$