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求解一道不定积分高数题.∫Cos^6x/sin^2x Dx =?附...

arctan(tanx)=x

亲 麻烦点一下采纳谢谢

(sinx*cosx)^2=0.25*sin(2x)^2 积分=-2/sin(2*x)*cos(2*x)+C

1.将分母变为sin2x即原式为∫[(4cos2x/sin^2(2x))]dx 2.进行换元即2x变为t,原式变为∫[(2cos2x/sin^2t)]dt. 3继续换元,可观察到(sin t)'=cost.所以原式等于2∫[(1/sin^2t]d(sint). 4.得出答案为:(-2/sint)+c 5.将t换回为2x有(-2/sin...

还需要帮忙的话可以先采纳再详解

∫sin(2x+1)dx = 1/2∫sin(2x+1)d(2x+1) = -1/2*cos(2x+1)+C

∫cos2xdx =∫cos2xd(1/2)2x =-1/2sin2x+C

先用积化和差公式变为简单三角函数,再用凑微分法计算。

你好 ∫x^2sin2xdx =-1/2∫x^2d(cos2x) =-1/2[cos2x*x^2-∫2x*cos2xdx] =-1/2[cos2x*x^2-∫xd(sin2x)] =-1/2[cos2x*x^2-(sin2x*x-∫sin2xdx)] =-1/2cos2x*x^2+1/2sin2x*x-1/2∫sin2xdx =-1/2cos2x*x^2+1/2sin2x*x+1/4cos2x+C 【数学辅导团】为您...

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