推 gggg9999 :感謝 不過這好像很程設哦@@ 12/18 01:32
※ 引述《gggg9999 (居九)》之銘言：
: ∫(e^5x)(5/x^2-2/x^3)dx=?
原式 = 5∫(e^5x)(1/x^2)dx - 2∫(e^5x)(1/x^3)dx
consider I_k = ∫(e^5x)(x^-k)dx, then
let f = e^5x f' = 5*e^5x
g' = x^-k g = (1/-k+1)x^(-k+1)
by part
I_k = (e^5x) * [1/-(k-1)] * x^[-(k-1)] - [1/-(k-1)]∫(e^5x)[x^-(k-1)]dx
= (e^5x) * [1/-(k-1)] * x^[-(k-1)] - [1/-(k-1)] * I_(k-1)
照這個遞迴關係就可以解出來了
其實根本不用這麼複雜啦…就 by part 慢慢解就可以了XD
: ∫-cos(2x)cot(2x)dx=?
-∫cos(2x)*cos(2x) / sin(2x) dx
= (1/2)∫[cos(2x)/sin(2x)]^2 * 2(-sin2x) dx = S
let u = cos 2x, then du = -2 sin2x dx
=> S = (1/2)∫u^2 / (1-u^2) du
= (1/2)[∫(-1)du + ∫1/(1-u^2) du]
= (1/2)[ -u + arctanh u ]
= (1/2)[ -cos 2x + arctanh(cos 2x)]
有請 Wolfram alpha 幫忙驗算
http://goo.gl/3mqdX
另外其實第二題也可以看成是 R(cos x,sin x)
即由 cos,sin組成的 rational function
consider t = tan(x/2)
=> cosx = (1-t^2)/(1+t^2) sinx = (2t)/(1+t^2)
x = 2 arctan x , dx = [2/(1+t^2)] dt
也就是說可以把其轉換成一堆 t 的分式
同理可應用在 R(sinh,cosh)
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