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求不定积分∫sin^4xCos^3xDx

=∫sin^4x(1-sin^2x)dsinx上面就是将sinx作为自变量,你可设sinx=u则:=fu^4(1-u^2)du=f[u^4-u^6]du 公式:(u^n)'=(n-1)^(n-1) ; fu^ndu=1/(n+1) *u^(n+1)+c=fu^4du-fu^6du=1/5u^5-1/7u^7+c再将u=sinx代入=1/5sin^5x-1/7sin^7x+c

前面是sinx的4次方还是sin4x啊

∫sin4xcosxdx =½∫(sin5x+sin3x)dx =-½·[(1/5)cos5x+⅓cos3x]+C =-(3cos5x+5cos3x)/30 +C

∫ (sin^4x)*(cos^2x) dx=1/16*x-1/64*sin4x-1/48*(sin2x)^3+C 解:∫ (sin^4x)*(cos^2x) dx =∫ ((1-cos2x)/2)^2*((cos2x+1)/2) dx =1/8∫ (1-cos2x))^2*(1+cos2x) dx =1/8∫ (1-cos2x-(cos2x)^2+(cos2x)^3) dx =1/8∫ 1 dx-1/8∫ cos2x dx-1/8∫ (cos2...

∫xcos^4(x/2)/sin^3(x)dx的结果为-x/(8*sin^2(x/2))-cot(x/2)/4+C。 解:∫(xcos^4(x/2))/sin^3(x)dx =∫(xcos^4(x/2))/(2sin(x/2)cos(x/2))^3dx =∫(xcos^4(x/2))/(8*sin^3(x/2)cos^3(x/2))dx =1/8∫(xcos(x/2))/(sin^3(x/2))dx =1/4∫x/(sin^3(x/2)...

如图所示、

∫SIN^4XCOS^4XDX =∫sin^2(2x)/4dx =∫(sin^2(2x)-1/2+1/2)/4dx =∫(1-(1-2sin^2(2x))/8dx =∫(1-cos(4x))/8dx =x/8-sin(4x)/32+C

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