Question

Evaluate:
${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}si{n}^{7}xdx$

Answer 1

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{7}xdx$

$=2{\int }_0^{\frac{\pi }{2}}{\sin }^{6}x\sin xdx$

$=2{\int }_0^{\frac{\pi }{2}}{(1-{\cos }^{2}x)}^3\sin xdx$

Let $t=\cos x\Rightarrow dt=-\sin xdx$

When $x=0\Rightarrow t=1$

$x=\frac{\pi }{2}\Rightarrow t=0$

$=-2{\int }_1^0{(1-t^2)}^{3}dt$

$=2{\int }_0^1{(1-t^2)}^{3}dt$

$=2{\int }_0^1(1-t^6-3t^2+3t^4)dt$

$=2{[t-\frac{t^7}{7}-\frac{3t^3}{3}+\frac{3t^5}{5}]}_0^1$

$=2[1-\frac{1}{7}-\frac{3}{3}+\frac{3}{5}-0]$

$=2[-\frac{1}{7}+\frac{3}{5}]$

$=2\left(\frac{-5+21}{20}\right)=2\times \frac{16}{20}=\frac{8}{5}$