Evaluate the integral `int_0^(pi/2) sin^5x dx`

Saturday, November 29, 2014

The integral
$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\sin}}^{{5}}{x}{\left.{d}{x}\right.}$ has to be determined.

$\displaystyle\int{{\sin}}^{{5}}{x}{\left.{d}{x}\right.}$

=> $\displaystyle\int{\sin{{x}}}\cdot{{\sin}}^{{4}}{x}{\left.{d}{x}\right.}$

=> $\displaystyle\int{\sin{{x}}}\cdot{{\left({1}-{{\cos}}^{{2}}{x}\right)}}^{{2}}$
dx

let $\displaystyle{\cos{{x}}}={y}$ , $\displaystyle{\left.{d}{y}\right.}=-{\sin{{x}}}{\left.{d}{x}\right.}$

=> $\displaystyle-\int{\left({1}\right.}$
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