☞有图☜我大一，请问下这四道求不定积分的题目怎么做？过程。。

Question

如题所述

Answer 1

4. (sinx)^4
= (sinx^2)^2
= ((1 - cos2x)/2)^2
= (1 - 2cos2x + (cos2x)^2)/4
= 0.25 - 0.5cos2x + 0.125(1 + cos4x)
= (cos4x)/8 - (cos2x)/2 + 3/8
∫ (sinx)^4dx
= ∫ ((cos4x)/8 - (cos2x)/2 + 3/8)dx
= ∫ ((cos4x)/8)dx - ∫ ((cos2x)/2)dx + ∫ (3/8)dx
= (1/32)∫ cos4xd4x - (1/4)∫ cos2xd2x + (3x/8)
= (sin4x)/32 - (sin2x)/4 + (3x/8) + C追答

3. ∫ √xsin√x dx
令y = √x,dy = 1/2√x dx => dx = 2y dy
= ∫ ysiny • 2y dy = 2∫ y²siny dy = -2∫ y² dcosy
= -2y²cosy + 2∫ cosy dy²,分部积分法
= -2y²cosy + 4∫ ycosy dy
= -2y²cosy + 4∫ y dsiny
= -2y²cosy + 4ysiny - 4∫ siny dy,分部积分法
= -2y²cosy + 4ysiny + 4cosy + C
= -2xcos√x + 4√xsin√x + 4cos√x + C,回代x = y²

1.1/3e^3x+2

追问

您可以写在纸上吗😂

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