数学　積分　2018/5-2012/2　Yuji.W

☆ 積分 1/(x^2+A^2)^n ☆

◎ 1/root(x^2+A^2)　1/(x^2+A^2)^n

〓　${dx/r^3}　〓　

■ r=root(x^2+A^2)　のとき、

　r;x=(1/2)*(2*x)/r=x/r　�@

　(1/r);x=[(1/r);r]*[r;x]=(-1/r^2)*(x/r)=-x/r^3　�A

　(x/r);x
=(x;x)/r+x*[(1/r);r]
=1/r-x*(x/r^3)
=1/r-x^2/r^3
=(r^2-x^2)/r^3
=A^2/r^3　�B

�@より　${(x/r)*dx}=r ★_

�Aより　${(x/r^3)*dx}=-1/r ★_

�Bより　${(1/r^3)*dx}=(1/A^2)*x/r ★_

〓　積分 1/(x^2+1)^n　〓　

◎ tan を利用する

● 1+tan(a)^2=1+[sin(a)/cos(a)]^2=1/cos(a)^2

　tan(a);a=[sin(a)/cos(a)];a=1/cos(a)^2=1+tan(a)^2

◆ x=tan(a)〔 -Pi/2<a<Pi/2 〕と置くと、

　dx=[1+tan(a)^2]*da=(x^2+1)*da　[1/(x^2+1)]*dx=da

　[1/root(x^2+1)]=cos(a)

■ ${[1/(x^2+1)]*dx}=${da}=a ★_

■ [1/(x^2+1)^(3/2)]*dx=[1/root(x^2+1)]*da=cos(a)*da

　${[1/(x^2+1)^(3/2)]*dx}=${cos(a)*da}=sin(a) ★_

〓　積分 1/(x^2+A^2)^n　〓　

◆ A>0　　x^2+A^2=A^2*[(x/A)^2+1]

まず　x/A=h　と置くと　x^2+A^2=A^2*(h^2+1)　dx=A*dh

さらに　h=tan(a)〔 -Pi/2<a<Pi/2 〕と置く　x=A*tan(a)

■ ${[1/(x^2+A^2)]*dx}
=(1/A^2)*${[1/(h^2+1)]*A*dh}
=(1/A)*${[1/(h^2+1)]*dh}
=a/A

≫　${[1/(x^2+A^2)]*dx}=a/A=arctan(x/A)/A ★_

■ [1/(x^2+A)^(3/2)]*dx=(1/A^3)/(h^2+1)^(3/2)

　${[1/(x^2+A)^(3/2)]*dx}
=(1/A^3)*${[1/(h^2+1)^(3/2)]*A*dh}
=(1/A^2)*${[1/(h^2+1)^(3/2)]*dh}
=sin(a)/A^2

≫　${[1/(x^2+A)^(3/2)]*dx}=sin(a)/A^2 ★_

{まとまってきたなあ!2017/5}

〓　{別解}積分 1/(x^2+A^2)^n　〓　

◆ x=A*tan(a)〔 A>0　-Pi/2<a<Pi/2 〕と置くと、

　x^2+A^2=A^2*[tan(a)^2+1]=A^2/cos(a)^2

　dx=A*[1+tan(a)^2]*da=[A/cos(a)^2]*da=(1/A)*(x^2+A^2)*da

　[1/(x^2+A^2)]*dx=(1/A)*da

■ ${[1/(x^2+A^2)]*dx}=(1/A)*${da}=a/A ★_

■ [1/(x^2+A)^(3/2)]*dx
=[1/root(x^2+A)]*(1/A)*da
=[cos(a)/A]*(1/A)*da
=[cos(a)/A^2]*da

　${[1/(x^2+A^2)^(3/2)]*dx}=(1/A^2)*${cos(a)*da}=sin(a)/A^2 ★_

〓　積分 1/(x^2+A^2)^n　〓　

◎ tan を利用する

● tan(a);a=1/cos(a)^2

■ x=A*tan(a)　と置く　x;a=A/cos(a)^2

[x:0~∞]=[a:0~Pi/2]　[x:-∞~∞]=[a:-Pi/2~Pi/2]

　x^2+A^2=A^2*[tan(a)^2+1]=A^2/cos(a)^2

　x^2+A^2=A^2*[tan(a)^2+1]=A^2/cos(a)^2

■ ${dx/(x^2+A^2)}
=${[A*da/cos(a)^2]/[A^2/cos(a)^2]}
=(1/A)*${da}
=a/A
=(1/A)*arctan(x/A) ★.

■ ${dx/(x^2+A^2)^(3/2)}
=${[cos(a)/A]^3*[A/cos(a)^2]*da}
=${[cos(a)/A^2]*da}
=sin(a)/A^2
=(1/A^2)*[x/root(x^2+A^2)] ★.

■ ${[1/(x^2+A^2)^2]*dx}
=${[cos(a)^2/A^2]^2*(A/cos(a)^2)*da}
=(1/A^3)*${cos(a)^2*da}
=(1/A^3)*${[(1+cos(2*a))/2]*da}
=[1/(4*A^3)]*[2*a+sin(2*a)]

■ ${[1/(A^2+x^2)^(5/2)]*dx}
=${(cos(a)/A)^5*(A/cos(a)^2)*da}
=(1/A^4)*${cos(a)^3*da}
=(1/A^4)*${(1-sin(a)^2)*cos(a)*da}
=(1/A^4)*${(1-sin(a)^2)*d(sin(a))}
=[1/(3*A^4)]*[3*sin(a)-sin(a)^3]

■ ${[x^2/(A^2+x^2)^(5/2)]*dx}
=${(A*sin(a)/cos(a))^2*(cos(a)/A)^5*(A/cos(a)^2)*da}
=(1/A^2)*${sin(a)^2*cos(a)*da}
=(1/A^2)*${sin(a)^2*d(sin(a))}
=[1/(3*A^2)]*sin(a)^3

{難しそうでも、トライしてみると、できるものだ!2014/2}

〓　双曲線関数の利用　〓