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∫x∧4sinxdx

来源:baiyundou.net   日期:2024-08-23

翟虾皇935不定积分∫sin5xdx=______. -
蒯鸦刷17771109370 ______[答案] 由题意 不定积分∫sin5xdx=- 1 5cos5x+C 故答案为;- 1 5cos5x+C.

翟虾皇935∫sinxdsinx=多少 -
蒯鸦刷17771109370 ______ 令sinx=y ,则∫sinxdsinx=∫ydy= (y^2)/2 所以原式=〔(sinx)^2〕/2=(1-cos2x)/4

翟虾皇935不定积分 ∫e∧x sinx dx 等于 -
蒯鸦刷17771109370 ______[答案] ∫e^xsinxdx=∫sinxde^x=e^xsinx-∫e^xdsinx=e^xsinx-∫e^xcosxdx=e^xsinx-∫cosxde^x=e^xsinx-e^xcosx+∫e^xdcosx=e^xsinx-e^xcosx-∫e^xsinxdx所以∫e^xsinxdx=(e^xsinx-e^xcosx)/2+C

翟虾皇935∫sin3xcosxdx -
蒯鸦刷17771109370 ______[答案] sin3xcosx积化和差 =(sin4x-sin2x)/2 所以∫sin3xcosxdx =(1/2)∫(sin4x-sin2x)dx =(1/2)*[∫sin4xdx-∫sin2xdx] =(1/2)*[(1/4)*∫sin4xd4x-(1/2)∫sin2xd2x] =-(1/8)cos4x+(1/4)*cos2x+C

翟虾皇935∫sin√xdx如何求解这个积分?∫sin√xdx怎么求解, -
蒯鸦刷17771109370 ______[答案] 令t=√x, ∫sin√xdx =∫sintdt^2 =2∫tsintdt =-2∫tdcost =-2(tcost-∫costdt)(分部积分) =-2(tcost-sint) =-2(√xcos√x-sin√x)

翟虾皇935∫sinx/e∧xdx的过程和结果(e∧x) -
蒯鸦刷17771109370 ______[答案] ∫ sinx dx / e^x= -∫ dcosx / e^x= -{cosx/e^x - ∫ cosx d(1/ e^x)}= -cosx/e^x + ∫ cosx [-e^x/e^2x]dx}= -cosx/e^x - ∫ d(sinx) /e^x= -cosx/e^x - {sinx/e^x - ∫sinx/e^x dx}= -cosx/e^x - sinx/e^x - ∫si...

翟虾皇935∫sinx∧2dx=? -
蒯鸦刷17771109370 ______ ∫sin²xdx=∫{(1-cos2x)/2}dx =x/2-1/4∫cos2xd2x =x/2-sin2x/4+C 其中C为任意常数.

翟虾皇935∫1x2sin1xdx. -
蒯鸦刷17771109370 ______[答案] 原式=−∫sin 1 xd 1 x(4分)=cos 1 x+C 利用凑微分计算可得, ∫ 1 x2sin 1 xdx =−∫sin 1 xd( 1 x) =cos 1 x+C.

翟虾皇935∫cos3xsinxdx -
蒯鸦刷17771109370 ______[答案] ∫cos3xsinx dx,用积化和差公式 =(1/2)∫(sin4x-sin2x) dx =(1/2)∫sin4xdx-(1/2)∫sin2xdx =(1/2)·(1/4)(-cos4x)-(1/2)·(1/2)(-cos2x)+C =(1/4)cos2x-(1/8)cos4x+C

翟虾皇935∫sin3xsin5xdx怎么求 -
蒯鸦刷17771109370 ______[答案] 须知公式:sinAsinB = (1/2)[cos(A-B)-cos(A+B] ∫sin(3x)sin(5x) dx = (1/2)∫[cos(-2x)-cos(8x)] dx = (1/2)∫cos2x dx - (1/2)∫cos8x dx = (1/4)∫cos2x d(2x) - (1/16)∫cos8x d(8x) = (1/4)sin2x - (1/16)sin8x + C

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