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2CosxCos2xDx

∫(-π/2~π/2)[cosx-2cosx(sinx)^2 ]dx=∫(-π/2~π/2)coxdx-∫(-π/2~π/2)2cosx(sinx)^2dx=sinx(-π/2~π/2)-2∫(-π/2~π/2)(sinx)^2dsinx=2-2/3[(sinx)^3](-π/2~π/2)=2-4/3=2/3

∫[0→π/2] 2xcos2x dx =∫[0→π/2] x d(sin2x) =xsin2x - ∫[0→π/2] sin2x dx =xsin2x + (1/2)cos2x |[0→π/2] =-(1/2) - (1/2) =-1 希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢。

∫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

你写的式子感觉都不对,有歧义 1.猜测你想表达的意思是: ∫ x cos(2 x) dx = 1/2 x sin(2 x)-1/2 ∫ sin(2 x) dx 令 u = 2 x 则 du = 2 dx: = 1/2 x sin(2 x)-1/4 ∫ sin(u) du = (cos(u))/4+1/2 x sin(2 x)+C 代回 u = 2 x: = 1/4 cos(2 x)+x sin...

cos2x=cos²x-sin²x=2cos²-1=1-2sin²x=(1-tan²x)/(1+tan²x)

两次分部积分∫x??cos2xdx=(1/2)∫x??d(sin2x)=(1/2)[x??sin2x-∫sin2xd(x??)]=(1/2)x??sin2x-∫xsin2xdx=(1/2)x??sin2x+(1/2)∫xd(cos2x)=(1/2)x??sin2x+(1/2)[xcos2x-∫cos2xdx]=(1/2)x??sin2x+(1/2)xcos2x-(1/2)∫cos2xdx=(1/2)x??sin2x+(1/2)xcos2x...

应该是∫(sinx)^2cos2xdx,用降幂公式把原式打开即可,解法如下:

有一个地方打错了,在“证明”后面,是i是虚数单位,不是n.

cos2x=cos²x-sin²x=2cos²x-1=1-2sin²x 所以1-cos2x=1-(1-2sin²x)=2sin²x 原式=∫x/2sin²x dx =1/2*∫x/sin²xdx =1/2*∫xcsc²xdx =-1/2*∫xdcotx =-1/2*xcotx+1/2*∫cotxdx =-1/2xcotx+1/2∫cosx/sinxdx ...

用两次分部积分,结果等于π/2(见图片)

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