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x*[(sinx)^2]的不定积分

∫x/(sinx)^2dx=∫x(cscx)^2dx=-∫xdcotx=-xcotx+∫cotxdx=-xcotx+∫cosx/sinxdx=-xcotx+lnsinx+C

你好!∫ x / (sinx)^2 dx= - ∫ x dcotx= - xcotx + ∫ cotx dx= - xcotx + ln|sinx| +C 满意请好评o(∩_∩)o

x(sinx)^2=x/2(1-cos2x) 然后用分步积分法就可以了

- cosx + c

∫x/(sinx)^2dx=-xcotx+∫cotxdx=-xcotx+In|sinx|+C

∫ (x-sinx)^2dx =∫ (x^2 - 2xsinx - sinx^2)dx =x^3/3 - ∫2xsinxdx - ∫sinx^2dx =x^3/3 - 2xcosx + 2∫cosxdx - ∫(1-cos2x)/2dx =x^3/3 - 2xcosx + 2sinx - x/2 + sin2x/4 + c

∫(x-sinx)^2*sinxdx=∫(x^2-2x*sinx+sin^2x)*sinxdx=-∫xdcosx-2∫ xsinx dx+∫(1-cos^2x)sinxdx=-xcosx+2xsinx+cosx+cos^3x/3+1/2(sin2x-2xcos2x)+c

∫xsinxdx=1/2∫x(1-cos2x)dx =x/6-(1/4)∫xd(sin2x) =x/6-(1/4)xsin2x+(1/2)∫xsn2xdx =x/6-(1/4)xsin2x-(1/4)∫xd(cos2x) =x/6-(1/4)xsin2x-(1/4)xcos2x+(1/8)∫cos2xd(2x) =x/6-(1/4)xsin2x-(1/4)xcos2x+(1/8)sin2x+c.

分部积分 ∫e^xsinxdx=∫sinxde^x=sinx*e^x-∫e^xdsinx=sinx*e^x-∫e^xcosxdx=sinx*e^x-∫cosxde^x=sinx*e^x-cosx*e^x+∫e^xdcosx=sinx*e^x-cosx*e^x-∫e^xsinxdx 所以2∫e^xsinxdx=sinx*e^x-cosx*e^x 所以∫e^xsinxdx=e^x(sinx-cosx)/2

∫f(sinx)cosxdx =∫f(sinx)dsinx 因为∫f(x)dx=1/(1+x^2) +c 所以∫f(sinx)dsinx = 1/[1+(sinx)^2] +c 那么∫f(sinx)cosxdx =1/[1+(sinx)^2] +c 哪里看不懂 发消息问我 乐意解答.

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