数学 積分

質問

問16の問題の時がよく分かりません。問題集の問題なのですが、答えも解き方も略になっていて分かりません。出来れば(1)(2)(3)の解き方と答えを教えて下さい。お願いします。

回答No.3

info222_ · ベストアンサー

(続く)
問16
(3)

ルート内>=0, 分母≠0 より
x<-2, x>2
f(x)=1/(x^4*(x^2-4)^(1/2))

g1(x)=(x^2-4)^(1/2)/x, g2(x)=(x^2-4)^(1/2)/x^3
g1'(x)=(x^2-4)^(1/2)(1/x)'+((x^2-4)^(1/2))'(1//x)
=-(1/x^2)(x^2-4)^(1/2)+(1//x)x/(x^2-4)^(1/2)
=1/(x^2-4)^(1/2)-(1/x^2)(x^2-4)^(1/2)
g2'(x)=(x^2-4)^(1/2)(1/x^3)'+((x^2-4)^(1/2))'(1//x^3)
=-(3/x^4)(x^2-4)^(1/2)+(1/x^3)x/(x^2-4)^(1/2)
=1/(x^2(x^2-4)^(1/2))-(3/x^4)(x^2-4)^(1/2)
a g1'(x)+b g2'(x)=f(x)
となるa,bを求めると
a=1/24, b=1/12

I=∫ f(x) dx=∫ (a g1 ' (x)+ b g2 ' (x)) dx=a g1(x)+b g2(x) +C
=(1/24)(1/x)(x^2-4)^(1/2) +(1/12)(1/x^3)(x^2-4)^(1/2) +C (C=任意定数)
=(x^2+2)(√(x^2-4))/(24x^3)+C (C=任意定数) ... (Ans.)

回答No.2

info222_

16.
(2)
x=a*sinh(t)
x^2+a^2=a^2*cosh^2(t)
a*cosh(t)=(x^2+a^2)^(1/2)
tanh(t)=x/(x^2+a^2)^(1/2)
dx=a*cosh(t) dt
I=∫x^2/(x^2+a^2)^(7/2) dx=∫ a sinh(t)tanh(t)/(a^5cosh^5(t)) dt
=(1/a^4)∫ sinh^2(t)/cosh^6(t) dt
=(1/a^4)∫ tanh^2(t)(1-tanh^2(t)) (tanh(t))' dt
=(1/a^4) {(1/3)tanh^3(t)-(1/5)tanh^5(t)} +C
=(1/a^4){(1/3)x^3/(x^2+a^2)^(3/2)-(1/5)x^5/(x^2+a^2)^(5/2)} +C
=(x^3/a^4){(1/3)(x^2+a^2)-(1/5)x^2}/(x^2+a^2)^(5/2) +C
=(1/(15a^4)) x^3 (2x^2+5a^2)/(x^2+a^2)^(5/2) +C (C=任意定数) ... (Ams.)

(とりあえず(2)まで)

回答No.1

info222_

問16
(1)
-a<=x<=a
x=a*sin(t) (-pi/2<=t<=pi/2)
tan(t)=sin(t)/cos(t)=x/(a^2-x^2)^(1/2)
dx=a*cos(t) dt
I=∫1/(a^2-x^2)^(5/2) dx=∫a*cos(t)dt/(a^5 (|cos(t)|)^5)
=(1/a^4) ∫ dt/(cos(t))^4
=(1/a^4) ∫ (1+tan^2(t))(tan(t))' dt
=(1/a^4) (tan(t)+tan^3(t)/3) +C(C:任意定数)
=(1/a^4) (x/(a^2-x^2)^(1/2)+x^3/(a^2-x^2)^(3/2)/3) +C
=(x/(3a^4)) (3a^2-2x^2)/(a^2-x^2)^(3/2)+C(C:任意定数) ... (Ans.)

(とりあえず(1)だけ)