15题不定积分，需要详解？ 高等数学，求不定积分。第15题怎么做？求详细过程！
15\u9898\uff0c\u6c42\u4e0d\u5b9a\u79ef\u5206\u3002\u6c42\u8be6\u89e3\u6b65\u9aa4\u3002\u5206\u90e8\u7684\u5177\u4f53\u505a\u6cd5=-(1/2)∫√(1-x^2) d(1-x^2)
=-(1/2)[ (2/3)(1-x^2)^(3/2)] +C
=-(1/3)(1-x^2)^(3/2) +C
∫x√(1-x²)dx
=-1/2∫√(1-x²)d(1-x²)
=-1/2×2/3∫d[(1-x²)√(1-x²)]
=-1/3(1-x²)√(1-x²)+ C
2。∫xe^(2x)dx=(1/2)∫xde^(2x)=(1/2)[xe^(2x)-∫e^(2x)dx]=(1/2)[xe^(2x)-(1/2)e^(2x)]+C
=(1/2)[x-(1/2)]e^(2x)+C
6。∫x²(e^x)dx=∫x²d(e^x)=x²e^x-2∫x(e^x)dx=x²e^x-2∫xd(e^x)=x²e^x-2[xe^x-∫(e^x)dx]
=x²e^x-2[xe^x-(e^x)]+C=(x²-2x+2)e^x+C
9。∫x²sinxdx=-∫x²d(cosx)=-[x²cosx-2∫xcosxdx]=-[x²cosx-2∫xd(sinx)]=-[x²cosx-2xsinx+∫sinxdx]
=-[x²cosx-2xsinx-cosx]+C=-x²cosx+2xsinx+cosx+C
15。∫(e^x)cosxdx=∫(e^x)d(sinx)=(e^x)sinx-∫sinxe^xdx=(e^x)sinx+∫(e^x)dcosx
=(e^x)sinx+(e^x)cosx-∫cosx(e^x)dx
移项得∫(e^x)cosxdx=(1/2)(sinx+cosx)e^x+C.
绛旓細浠1-x=t锛屽垯x=1-t锛宒x=-dt 鍘熷紡=-鈭玹^4 dt =-t^5 /5+C =-(1-x)^5 /5 +C =(x-1)^5 /5 +C
绛旓細c
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绛旓細绗竴棰 鈭玞os鈭歺dx=鈭2tcostdt=2tsint锛嶁埆2sintdt 绗簩棰 鈭玡^鈭歺dx=鈭2te^tdt=2te^t锛嶁埆2e^tdt
绛旓細鈭(dx/(4+9x^2 ))=(1/6)鈭玠(3x/2)/((3x/2)²+1)=(1/6)arctan(3x/2)+C (arctanx)'=1/(x²+1)鈭玠x/(x²+1)=arctanx+C 閫氳繃鎹㈠厓锛屼护t=3x/2鍗冲彲
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绛旓細浠=sint (-蟺/2鈮鈮は/2)鍒檇x=costdt 鍘熷紡=鈭玞ostdt/(1+cost)锛濃埆[1-1/(1+cost)]dt 锛濃埆{1-(1/2)[sec(t/2)]^2}dt 锛漷锛峵an(t/2)锛婥 锛漷锛峓sint/(1+cost)]锛婥 锛漚rcsinx锛峹/[1+鈭(1-x^2)]锛婥
绛旓細鈭玞os^2 3tdt =鈭紙1+cos6t锛/2dt =鈭1/2+cos6t/2dt =1/2t+1/12sin6t+C 鈭玠x/sinxcosx =鈭2(dx)/[2sinxcosx]=鈭玔2dx]/[tanxcos²x]=2鈭玔dtanx]/tanx =2ln鈹倀anx鈹+C 鍙︿竴绉嶆柟娉曟槸 2鈭玞scx dx=2ln鈹俢scx-cotx鈹+C 3.鈭玠x/e^2x+e^-2x 浠^x=t,鍒檟=lnt....
