∫ sin(4x)cos(4x) dx that's how I did it.
U=sin4x
1/4Du= cos4x
1/4∫udu
1/4*1/2(u)^2+c
= 1/8(sin(4x))^2+c
And then I seen it worked out as
∫ sin(4x)cos(4x) d
U=4x
1/4du=4dx
1/4∫sin(u)cos(u)du ---> now you use the sin(2u) identity. I know how to finish it but which work is right ???
Calculus help. Intergration U-sub?
2014-01-25 5:06 pm
回答 (3)
2014-01-25 5:12 pm
✔ 最佳答案
it's a lot easier if you do it this way. 2 sin 4x cos 4x = sin 8x
so int [ ( 1 / 2 ) sin 8x dx ]
= ( 1 / 2 ) * int [ sin 8x dx ]
= ( 1 / 2 ) ( - 1 / 8 ) cos 8x + C
= ( - 1 / 16 )cos 8x + C
參考: my brain
2014-01-25 5:34 pm
This link has a step by step solution to your problem:
http://www.symbolab.com/solver/integral-calculator/%5Cint%20%20sin(4x)cos(4x)dx
Hope this helps
http://www.symbolab.com/solver/integral-calculator/%5Cint%20%20sin(4x)cos(4x)dx
Hope this helps
2014-01-25 5:29 pm
∫ sin(4x)cos(4x) dx
Let u = 4x
du = 4 dx
dx = (1/4) du
∫ sin(4x)cos(4x) dx = (1/4) ∫ sin (u) cos(u) du
Let t = sin (u)
dt = cos(u) du
(1/4) ∫ sin (u) cos(u) du = (1/4) ∫ t dt
= (1/4)(1/2) t^2
= (1/8) t^2
replace t by sin (u)
= (1/8) sin^2 (u)
replace u by 4x
= (1/8) sin^2 (4x) + C
Let u = 4x
du = 4 dx
dx = (1/4) du
∫ sin(4x)cos(4x) dx = (1/4) ∫ sin (u) cos(u) du
Let t = sin (u)
dt = cos(u) du
(1/4) ∫ sin (u) cos(u) du = (1/4) ∫ t dt
= (1/4)(1/2) t^2
= (1/8) t^2
replace t by sin (u)
= (1/8) sin^2 (u)
replace u by 4x
= (1/8) sin^2 (4x) + C
收錄日期: 2021-04-20 20:27:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140125090651AAV5Ld0