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∫xCos²xDx

∫xcos²xdx=∫x(1+cos2x)/2dx=1/2(∫xdx+∫xcos2xdx) =1/2(1/2x²+∫xcos2xdx) =1/2(1/2x²+1/2∫xdsin2x) =1/2(1/2x²+1/2(xsin2x-∫sin2xdx)) =1/2(1/2x²+1/2xsin2x+1/4cos2x)+C

作变量代换√(4-x)=t,就中以如图化简并求出这个不定积分。

cosx+3/xdx=1/12x24x254cosx+C ∫x212x22xsinx+3/12x2dx3cos3x-1/2xsinx-3/12x2dcos3x =-3/dx=1/4x212x23cosx34∫x218xsin3x+1/4x2cosx+3/4∫cosxdx24x2∫x2cos3x-1/4∫x23∫sinx32∫xcosxdx+1/sinx3sin3=-1/4x2 =-3/4∫x218xsin3x-1/18∫sin3xdx =-3/sin...

∫sin³xcos²xcos³xdx =∫sinx·(1-cos²x)cos²xcos³xdx =∫(-cos⁷x+cos⁵x)sinxdx =∫(cos⁷x-cos⁵x)d(cosx) =⅛cos⁸x -(1/6)cos⁶x +C

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设√x=t,则x=t²,dx=2tdt 原式=∫cos²t*2tdt=∫(2cos²t)*tdt =∫(cos2t+1)*tdt =∫cos2t*tdt+∫tdt =(1/2)∫td(sin2t)+(1/2)t² =(1/2)[t*sin2t-∫sin2tdt]+(1/2)t² =(1/2)tsin2t-(1/2)∫sin(2t)dt+(1/2)t² =(1/2)tsin2t+(...

∫sin²xcos⁵xdx =∫sin²x(1-sin²x)²·cosxdx =∫(sin⁶x-2sin⁴x+sin²x)d(sinx) =(1/7)sin⁷x -(2/5)sin⁵x+(1/3)sin³x +C =(15sin⁴x-42sin²x+35)sin³x/105 +C

是这个吗???

∫cos³xdx =∫cos²xd(sinx) =∫(1-sin²x)d(sinx) =sinx-1/3sin³x+C C为常数 行家正解,不明白可以追问!祝您学习进步 满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢

cos2x=2cos^2x-1 所以 1+cos2x=2cos²x

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