Use of substitution u=1/x
to find ∫[√(1-x^2)/x^4]dx
integration
2007-10-16 7:14 am
回答 (2)
2007-10-17 5:55 am
✔ 最佳答案
The integral is straightforward as follow:圖片參考：http://i187.photobucket.com/albums/x22/cshung/7007101504798.jpg
2007-10-16 21:59:29 補充：
Ooops... a little bit slower, remember to add the constant C.
2007-10-16 8:43 am
Use of substitution u=1/x
to find ∫[√(1-x^2)/x^4]dx
u=1/x;x=1/u
x^2=1/u^2;x^4=1/u^4
dx=(-1/u^2)du
∫[√(1-x^2)/x^4]dx
=∫[u^4√(1-1/u^2)](-1/u^2)du
=-∫{u^2√[(u^2-1)/u^2]}du
=-∫u√(u^2-1)du
=-(1/2)∫√(u^2-1)d(u^2-1)
=-(1/3)(u^2-1)^(3/2)+C
=-(1/3)(1/x^2-1)^(3/2)+C
to find ∫[√(1-x^2)/x^4]dx
u=1/x;x=1/u
x^2=1/u^2;x^4=1/u^4
dx=(-1/u^2)du
∫[√(1-x^2)/x^4]dx
=∫[u^4√(1-1/u^2)](-1/u^2)du
=-∫{u^2√[(u^2-1)/u^2]}du
=-∫u√(u^2-1)du
=-(1/2)∫√(u^2-1)d(u^2-1)
=-(1/3)(u^2-1)^(3/2)+C
=-(1/3)(1/x^2-1)^(3/2)+C
收錄日期: 2021-04-23 17:19:21
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