오일러 치환

이 항목의 스프링노트 원문주소

개요

\(R(x,\sqrt{ax^2+bx+c})\)의 적분
\(ax^2+bx+c=\frac{1}{a}\{(ax+b)^2+{ac-b^2}}\}\) 으로 쓴 다음
\(ac-b^2\)와 \(a\)의 부호에 따라, 적당히 치환

\(R(x,\sqrt{ax^2+bx+c})\) 형태의 적분을 유리함수의 적분으로 바꾸는 변수 \(x=x(t)\) 치환

\(y^2=ax^2+bx+c\)를 \(t\)에 대한 유리함수로 매개화하는 것이 가장 중요한 아이디어이다.
\(y-y_0 = t(x-x_0)\) passing through a point \((x_0,y_0)\)

제1오일러치환

\(a>0\) 일때, \(\sqrt{ax^2+bx+c}=t-\sqrt{a}x\) 로 치환
\(\int\sqrt{x^2-4}\,dx\)의 예
\(\sqrt{x^2-4}=t-x\)
\(x=\frac{4+t^2}{2t}\)
\(\int \frac{2t^4-16t^2+32}{8t^3}\,dt\)

multiply out.

제2오일러치환

\(c>0\) 일때, \(\sqrt{ax^2+bx+c}=xt+\sqrt{c}\) 로 치환
의 예

제3오일러치환

\(\int\sqrt{x^2-4}\,dx\)의 예
\(\sqrt{x^2-4}=t-x\)
\(x=\frac{4+t^2}{2t}\)
\(\int \frac{2t^4-16t^2+32}{8t^3}\,dt\)

The second Euler substitution: If the roots and of the quadratic polynomial are real, then

Since we can factor the polynomial and one root is 2, we can also use the 3. Euler substitution:

In the case when , that is, when (2) is a hyperbola, the first Euler substitution is obtained by taking \((x_0,y_0)\) as one of the points at infinity defined by the directions of the asymptotes of this hyperbola;

when the roots and of the quadratic polynomial \(ax^2+bx+c\) are real, the second Euler substitution is obtained by taking as \((x_0,y_0)\) one of the points or ;

finally, when , the third Euler substitution is obtained by taking as \((x_0,y_0)\) one of the points where the curve (2) intersects the ordinate axis, that is, one of the points .

http://www.integral-table.com/

http://books.google.com/books?id=E2IhMXPZMNIC&pg=PR8&lpg=PR8&dq=functions+with+elementary+integral+Analysis+by+Its+History&source=bl&ots=7GRnB0mT8k&sig=jpLHMzhVvPUFDTvIhCYojZWTYNo&hl=ko&ei=VU2HSuu2FpTOsQPMwInbAg&sa=X&oi=book_result&ct=result&resnum=3#v=onepage&q=&f=false

\(\int \sqrt{x^2+1}\,dx\)

http://www.goiit.com/posts/list/integration-euler-s-substitution-354.htm

http://pauli.uni-muenster.de/~munsteg/arnold.html