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求2½Cosx的原函数是多少

∫½sinx=-½cosx+C —————————————————— ∫dx/(2sinx) =½∫sinx/sin²xdx =-½∫d(cosx)/(1-cos²x) =-¼∫[1/(1-cosx)+1/(1+cosx)]d(cosx) =-¼[∫-1/(1-cosx)d(1-cosx)+∫1/(1+cosx)]d(1+cosx) =¼(ln|1-cosx|-...

解析: ∫(cosx)⁴dx = ∫(cos²x)²dx = ∫[(1+cos2x)/2]²dx =(1/4)∫[1+2cos2x+(cos2x)²]dx =(1/4)∫[1+2cos2x+(1+cos4x)/2]dx =(1/8)∫[2+4cos2x+(1+cos4x)]dx =(1/8)∫(3+4cos2x+cos4x)dx =(1/8)[3x+2sin2x+(1/4)sin4x]+C =...

x³sinx+3x²cosx-6sinx+C 解: 分部积分公式:∫udv=u*v-∫vdu ∫x³cosxdx =∫x³d(sinx) =x³sinx-∫sinxd(x³) =x³sinx-3∫x²sinxdx =x³sinx+3∫x²d(cosx) =x³sinx+3x²cosx-3∫cosxd(x²...

令arccosx=t,则x=cost ∫(arccosx)³dx =∫t³d(cost) =t³·cost-∫costd(t³) =t³·cost-3∫t²d(sint) =t³·cost-3t²sint+3∫sint(dt²) =t³·cost-3t²sint-6∫td(cost) =t³·cost-3t²sint...

∫dx/(cos²x+1) 分子分母同除以cos²x =∫sec²x/(1+sec²x)dx =∫d(tanx)/(1+1+tan²x) =√2/2∫d(tanx)/[1+(tanx)/√2)²] =√2/2·arctan[(tanx)/√2]+C

∫x²sinxdx=-∫x²dcosx=-x²cosx+∫cosxdx² =-x²cosx+∫2xdsinx=-x²cosx+2xsinx+2cosx+C

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