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∫xCos²xDx

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∫xcos^3 xdx =∫xcos^2 xcosxdx =∫xcos^2 xdsinx =∫x(1-sin^2 x)dsinx =∫xdsinx-∫xsin^2 x dsinx =xsinx-∫sinxdx-1/3∫xdsin^3 x =xsinx+∫dcosx-1/3(xsin^3 x -∫sin^3 x dx) =xsinx+cosx-xsin^3 x/3+1/3∫sin^2xsinxdx =xsinx+cosx-xsin^3 x/3-1/3∫...

定积分偶倍奇零 =2∫(0.π/2)sin²xcos²xdx =1/2∫sin²2xdx =1/4∫1-cos4xdx =x/4-sin4x/16 =π/8

(1) ∫x^5.√(1-x^2)dx let x=siny dx=cosy dy ∫x^5.√(1-x^2)dx =∫ (siny)^5 .(cosy)^2 dy =-∫ (1 -(cosy)^2)^2. (cosy)^2 dcosy =-∫[ (cosy)^2 - 2(cosy)^4 + (cosy)^6] dcosy =-(1/3)(cosy)^3 + (2/5)(cosy)^5 -(1/7)(cosy)^7 + C =-(1/3)(1-x^2...

答: 先证明柯西—布尼亚科夫斯基不等式:{∫(a到b)φ(x)ψ(x)dx}^2=0 所以本题不等式左边应用上式定理得: [∫(a到b)f(x)coskxdx]^2+[∫(a到b)f(x)sinkxdx]^2

1:[∫0 x cos^2tdt ]'= cos^2(X) 2:[∫x x^2 sin^2tdt ]'=2X sin^2 (X^2)-sin^2(X) 3[∫1 4 (X-1)^2/根号xdx]'=0 若计算积分即 ∫(1 4) (x - 1)²/√x dx= ∫(1 4) (x² - 2x + 1)/(x)^(1/2) dx= ∫(1 4) [x^(3/2) - 2(x)^(1/2)+ 1/(x)^...

=(1/3)x²sin3x+(2/9)xcos3x-(2/27)sin3x + c

(sinnπ+cosnπ-1)/nπ

如图

1/3 ∫xdsin3x 1/3 (∫sin3xdx-∫sin3x xdx) 1/3 (-(1/3)cos3x- -1/3∫xdcos3x) 1/3 (-(1/3)cos3x+ 1/3(∫cos3xdx-∫xcos3xdx)) 1/3 (-(1/3)cos3x+ 1/3(sin3x-∫xcos3xdx)) 1/9 ( sin3x-cos3x)-1/9 ∫xcos3xdx=∫xcos3xdx sin3x-cos3...

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