Question

一道微积分题目。

∫(sinx)^3(cosx)^5dx

Answer 1

∫(sinx)^3(cosx)^5dx
=∫(sin³x)(cos³x)cos²xdx
=∫(sinxcosx)³cos²xdx
=⅛∫(sin³2x)cos²xdx
=⅛∫(sin³2x)½(1+cos2x)dx
=⅛×½∫(sin³2x)dx + ⅛×½∫(sin³2x)cos2xdx
=⅛×½∫(sin²2x)sin2xdx + ⅛×¼∫(sin³2x)dsin2x
=-⅛×¼∫(sin²2x)dcos2x + ⅛×¼∫(sin³2x)dsin2x
=-⅛×¼∫(1-cos²2x)dcos2x + ⅛×¼∫(sin³2x)dsin2x
=-⅛×¼∫dcos2x + ⅛×¼∫cos²2xdcos2x + ⅛×¼∫(sin³2x)dsin2x
=-(1/32)cos2x + (1/96)cos³2x + (1/128)sin⁴2x + C

上面的结果是是用二倍角表达的，或改成原来的x角表达：

-(1/32)cos2x + (1/96)cos³2x + (1/128)sin⁴2x + C
=-(1/32)(1-2sin²x) + (1/96)(1-2sin²x)³ + (1/128)(2sinxcosx)⁴+ C
=(1/16)sin²x) + (1/96)(1-6sin²x+12sin⁴x-8(sinx)^6 + (1/8)sin⁴x cos⁴x + C
=(1/16)sin²x) + (1/96)(-6sin²x+12sin⁴x-8(sinx)^6 + (1/8)sin⁴x cos⁴x + C
=(1/16)sin²x) - (1/16)sin²x + (1/8)sin⁴x - (1/12)(sinx)^6 + (1/8)sin⁴x (1-sin²x)² + C
=(1/8)sin⁴x - (1/12)(sinx)^6 + (1/8)sin⁴x (1-sin²x)² + C
=(1/8)sin⁴x - (1/12)(sinx)^6 + (1/8)sin⁴x (1-2sin²x+sin⁴x) + C
=(1/8)sin⁴x - (1/12)(sinx)^6 + (1/8)sin⁴x - (1/4)(sinx)^6 + (1/8)(sinx)^8 + C
=(1/4)sin⁴x - (1/3)(sinx)^6 + (1/8)(sinx)^8 + C

Answer 2

∫(sinx)^3(cosx)^5dx=∫(cosx^2-1)(cosx)^5dcosx
=(cosx)^8/8-(cosx)^6/6+C

Answer 3

一个第二类间断点？

Answer 4

应该假设限制条件x≠-1，否则极限没有意义
|x|＜1时，f(x)＝1+x
|x|＞1时，f(x)＝0
x＝1时，f(x)＝1
所以x＝1是跳跃间断点，x＝-1是可去间断点

Answer 5

绝对值大于1的实数