My Math Forum integrating factor method

 Algebra Pre-Algebra and Basic Algebra Math Forum

 April 6th, 2009, 01:35 AM #1 Newbie   Joined: Apr 2009 Posts: 1 Thanks: 0 integrating factor method this is the end part of a question and i dont understand the final integrating part. now Y.IF = / Q.IF dX (/ = THE INTEGRATION SIGN) THEREFORE Y (1-X^2)^1/2 = / (1/1-X^2).(1-X^2)^1/2 dX Y.(1-X^2)^1/2 = / (1/(1-X^2)^1/2 dX / (1/(1-X^2)^1/2) dX = SIN^-1 X + C Y.(1-X^2)^1/2 = SIN^-1 X + C The bit that i do not get is how did they get from / (1/(1-x^2)^1/2) dX = sin^-1 x + C i cant see how they integrated to get to sin-1 x, is there a method that i do not know? thank you for helping!
 April 6th, 2009, 11:28 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,754 Thanks: 2137 $\text{Let }u\,=\,\sin^{-1}X,\text{ where }-\frac\pi2\le u\le\frac\pi2\text{, so that }X\,=\,\sin\,u\text{ and }dX\,=\,\cos\,u\,du\,=\,(1\,-\,X^2)^{1/2}du.$ $\int(1/(1\,-\,X^2)^{1/2}\,dX\,=\,\int du\,=\,u\,+\,C\,=\,\sin^{-1}X\,+\,C.$

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