`int_(0)^(pi) x sin^(2) xdx` का मान ज्ञात कीजिएः
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`int x. sin^(2) xdx =1/ int x(1-cos 2x)dx`
`=1/2 int (x-xcos 2x)dx`
`=1/4x^(2)-1/2int x cos 2 xdx`
`=1/4x^(2)-1/2 [(x sin 2x)/(2)-int (1. sin 2x)/(2)dx]`
`=1/4x^(2)-1/4x sin 2x-1/8 cos 2x`
`therefore underset(0)overset(pi)int x.sin^(2)xdx=1/4 [x^(2)-x sin 2x-1/2 cos 2x]_(0)^(pi)`
`=1/4 [pi^(2)-pisin 2pi-1/2 cos 2pi-0+0.sin 0+1/2 cos 0]`
`=1/4[pi-0+1/2+1/2]=(pi^(2))/(4)`