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求∫2xCos(x^2+1)Dx的不定积分.

凑微分即可

∫xcos2xdx =(1/2)∫xdsin2x =(1/2)x.sin2x -(1/2)∫sin2xdx =(1/2)x.sin2x +(1/4)cos2x + C

原式=1/2·∫(3+cos2x)/cos2x·dx =1/2·∫(3sec2x+1)dx =3/4·ln|sec2x+tan2x|+x/2+C 【基本公式】 ∫secxdx=ln|secx+tanx|+C

∫1/sin²xcos²x dx =∫1/sin²x dx+∫1/cos²x dx =-cotx + tanx + c =tanx-cotx + c

显然1+cos2x=2(cosx)^2 那么 原积分 =∫1/2(cosx)^2 dx =0.5 *∫1/(cosx)^2 dx =0.5tanx +C,C为常数

亲 麻烦点一下采纳谢谢

∫(x-1)cos2xdx =∫(xcos2x-cos2x)dx =∫xcos2xdx-∫cos2xdx =1/2∫xdsin2x-∫cos2xdx =1/2(xsin2x-∫sin2xdx)(分部积分法)-∫cos2xdx =1/2xsin2x-1/2∫sin2xdx-∫cos2xdx =1/2xsin2x-1/4∫sin2xd2x-∫cos2xdx =1/2xsin2x+1/4cos2x-1/2sin2x 你是第三步错了...

∫1/(1-cos2x) dx =∫1/[1-(1-2sin²x)]dx =∫1/2sin²x dx =(1/2)∫csc²xdx =-(1/2)cotx+c

∫1/(5cos^2x+3sin^2x)dx=∫1/(2cos^2x+3)dx=∫1/(cos2x+4)dx =1/15*根(15)*arctan(1/5*根(15)*sin(2*x)/(cos(2*x) + 1))

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