x*(sinx)^m在(0,π)的不定积分其中m为自然数
1个回答

x*(sinx)^m在(0,π)的【定积分】其中m为自然数:

令:x=π-t

∫(0,π) xsin^mxdx

=∫(π,0) (π-t)sin^m(π-t)d(π-t)

=∫(0,π) πsin^mtdt -∫(0,π) tsin^mtdt

= π∫(0,π) sin^mtdt -∫(0,π) xsin^mxdx

【此项左移】

= π/2∫(0,π) sin^mtdt

= π/2∫(0,π/2) sin^mtdt + π/2∫(π/2,π) sin^mtdt

令:x=π-t

= π/2∫(0,π/2) sin^mtdt + π/2∫(0,π/2) sin^mxdx

= π∫(0,π/2) sin^mxdx

① = π[(m-1)!/m!] m为奇数;

② = π²/2 [(m-1)!/m!] m为偶数.