∫x∧5（sinx）∧2dx

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∫x∧5（sinx）∧2dx
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∫x∧5（sinx）∧2dx
∫x∧5（sinx）∧2dx

∫x∧5（sinx）∧2dx
求不定积分∫x⁵sin²xdx
原式=(1/2)∫x⁵(1-cos2x)dx=(1/2)[∫x⁵dx-∫x⁵cos2xdx]=(1/2)[(1/6)x⁶-(1/2)∫x⁵dsin(2x)]
=(1/12)x⁶-(1/4)[x⁵sin2x-5∫x⁴sin2xdx]=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)∫x⁴d(cos2x)
=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)[x⁴cos2x-4∫x³cos2xdx]
=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)∫x³d(sin2x)
=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)[x³sin2x-3∫x²sin2xdx]
=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)∫x²d(cos2x)
=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)[x²cos2x-(1/2)∫xcos2xdx]
=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)x²cos2x-(15/32)∫xd(sin2x)
=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)x²cos2x-(15/32)[xsin2x-(1/2)∫sin2xd(2x)]
=(1/12)x⁶-(1/4)x⁵sin2x-(5/8)x⁴cos2x+(5/4)x³sin2x+(15/8)x²cos2x-(15/32)xsin2x-(15/64)cos2x+C

是定积分吧？高数里怎么会有这么麻烦的不定积分，没什么技巧，就是计算量大。

∫x∧5（sinx）∧2dx ∫x(sinx)∧5 dx 几道不定积分的做法1.∫(sinx)∧2*(cosx)∧5 dx 2.∫(1-cosx)/(x-sinx) dx 3.∫dx/(x*㏑x) 求不定积分∫（2x²+3e∧x+5sinx）dx ∫ arccos7x dx∫ xcos(2-x) dx∫ sinx/(5+3sinx) dx ∫(√sinx)cos∧5(x)dx ∫(sinx)/(cos∧2 x)dx怎么解? ∫sinX/1+X∧2dX怎么解 求∫1到5（｜2-x｜+｜sinx｜）dx ∫（上限5,下限1）(|2-x|+|sinx|)dx ∫(sinx-x^2)dx ∫x/(sinx)^2dx ∫ x/(sinx)^2dx 1.∫(sinx)^5(cosx)^3dx 2.∫(sinx)²(cosx)²dx 3.∫（e^x²+2x²e^x²）dx 求∫sinx/√（4-cos∧2x）dx 求三道不定积分∫√(1+sinx)dx,∫1/(x∧2+4x-5)dx,求三道不定积分∫√(1+sinx)dx,∫1/(x∧2+4x-5)dx,∫(3x+1)/(x∧2+4x-5)dx ∫(x/2+e^x+sinx)dx 设f（x）＝sinx∧2,则∫(1/x∧2)∫'（1/x）dx=