Question

${\int }_0^{\pi /2}{x}^2\sin xdx=\pi -k$.Find $k$ ?

Answer 1

${\int }_0^{\frac{\pi }{2}}{x}^{2}sin\left(x\right).dx$
$=-x^{2}cos\left(x\right)+\int 2x.cos\left(x\right).dx$
$=-x^{2}cos\left(x\right)+2x.sin\left(x\right)-\int 2sin\left(x\right).dx$
$=[-x^{2}cos(x)+2xsin(x)+2cos(x){]}_0^{\frac{\pi }{2}}$
$=\pi -2$
Hence
$k=2$.