Question

The value of the integral ${\int }_0^{\frac{\pi }{2}}si{n}^{5}xdx$ is

• A. $\frac{4}{15}$
• B. $\frac{8}{5}$
• C. $\frac{8}{15}$
• D. $\frac{4}{5}$

Answer 1

The correct option is C $\frac{8}{15}$

$I={\int }_0^{\frac{\pi }{2}}si{n}^{4}x.sinxdx$
$={\int }_0^{\frac{\pi }{2}}(1-co{s}^{2}x{)}^{2}sinxdx$
Put $cosx=t$
$\Rightarrow -sinxdx=dt$
$=-{\int }_1^0(1-t^2{)}^{2}dt={\int }_0^{(}{t}^4-2t^2+1)dt$
$[\because -{\int }_a^{b}f(x)dx={\int }_a^{b}f(x\left)dx\right]$
$=\frac{1}{5}(t^5{)}_0^1-\frac{2}{3}(t^3{)}_0^1+(t{)}_0^1$
$=\frac{1}{5}-\frac{2}{3}+1=\frac{3-10+15}{15}=\frac{8}{15}$