Question

高数求助 如何从第一步得到第二步的，谢谢

Answer 1

let
x=-u
dx=-du
u=-π/4, x=π/4
u=π/4, x=-π/4
(1/2)∫(-π/4->π/4) [1/(2^u +1) ] (cos2u)^4 du
=(1/2)∫(π/4->-π/4) [1/(2^(-x) +1) ] (cos2x)^4 [-dx]
=(1/2)∫(-π/4->π/4) [1/(2^(-x) +1) ] (cos2x)^4 dx
=(1/2)∫(-π/4->π/4) [ 2^x/(2^(x) +1) ] (cos2x)^4 dx

追问

然后怎么出来的1/4呢？

追答

I
=(1/2)∫(-π/4->π/4) [1/(2^x +1) ] (cos2x)^4 du
=(1/2)∫(-π/4->π/4) [ 2^x/(2^(x) +1) ] (cos2x)^4 dx
2I
=(1/2)∫(-π/4->π/4) [1/(2^x +1) ] (cos2x)^4 du
+(1/2)∫(-π/4->π/4) [ 2^x/(2^(x) +1) ] (cos2x)^4 dx
=(1/2)∫(-π/4->π/4) [ (1+2^x)/(2^x +1) ] (cos2x)^4 du
I=(1/4)∫(-π/4->π/4) [ (1+2^x)/(2^x +1) ] (cos2x)^4 du

追问

这一部分是为零的对吗？

所以在下一个等号中，两个式子就可以结合在一起了

追答

I
=(1/2)∫(-π/4->π/4) [1/(2^x +1) ] (cos2x)^4 du
=(1/2)∫(-π/4->π/4) [ 2^x/(2^(x) +1) ] (cos2x)^4 dx
2I
=(1/2)∫(-π/4->π/4) [1/(2^x +1) ] (cos2x)^4 du
+(1/2)∫(-π/4->π/4) [ 2^x/(2^(x) +1) ] (cos2x)^4 dx
=(1/2)∫(-π/4->π/4) [ (1+2^x)/(2^x +1) ] (cos2x)^4 du
I
=(1/4)∫(-π/4->π/4) [ (1+2^x)/(2^x +1) ] (cos2x)^4 du
=(1/4)∫(-π/4->π/4) (cos2x)^4 du
=(1/2)∫(0->π/4) (cos2x)^4 du
=(1/8)∫(0->π/4) (1+ cos4x)^2 du
=(1/8)∫(0->π/4) [1+ 2cos4x + (cos4x)^2 ] du
=(1/16)∫(0->π/4) (3+ 4cos4x + cos8x ) du
=(1/16) [3x+ sin4x + (1/8)sin8x]|(0->π/4)
=(1/16) (3π/4)
=3π/64

追问

谢谢😜

Answer 2

分布积分换元法。

追问

可以写一下详细过程嘛？