∫(2x-1)除以根号x dx ∫cosx dx +∫-2(sinx)^2 乘以cosx dx+∫(sinx)^4乘以cosx dx

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∫(2x-1)除以根号xdx∫cosxdx+∫-2(sinx)^2乘以cosxdx+∫(sinx)^4乘以cosxdx∫(2x-1)除以根号xdx∫cosxdx+∫-2(sinx)^2乘以cosxdx

∫(2x-1)除以根号x dx ∫cosx dx +∫-2(sinx)^2 乘以cosx dx+∫(sinx)^4乘以cosx dx
∫(2x-1)除以根号x dx ∫cosx dx +∫-2(sinx)^2 乘以cosx dx+∫(sinx)^4乘以cosx dx

∫(2x-1)除以根号x dx ∫cosx dx +∫-2(sinx)^2 乘以cosx dx+∫(sinx)^4乘以cosx dx
怎么全是不定积分啊?古怪
∫(2x-1) /√x dx ∫cosx dx +∫-2(sinx)^2 cosx dx+∫(sinx)^4 cosx dx
= [ ∫2√x dx-∫1/√x dx] ∫cosx dx -∫2(sinx)^2 d(sinx)+∫(sinx)^4 d(sinx)
=[(4/3) x^(3/2)+C1-2√x+C2](sinx+C3) - 2(sinx)^3 /3 + C3 + (sinx)^5 / 5 +C4
=[(4/3) x^(3/2) - 2√x + C](sinx + C') - 2(sinx)^3 /3 + (sinx)^5 / 5 +C"
其中C,C',C" 可以是任意常数