Wikipedia:Reference desk/Archives/Mathematics/2012 October 15

= October 15 =

Indefinite integral of inverse function
If we know the indefinite integral g(x) = $$\int f(x)dx,$$ can we find the indefinite integral $$\int f^{-1}(y)dy$$ directly from g(x) without performing an integration? Duoduoduo (talk) 23:20, 15 October 2012 (UTC)
 * Integration by parts shows that
 * $$\int_{f(a)}^{f(x)}f^{-1}(t)dt + \int_a^x f(s)ds = xf(x)-af(a).$$ From this you should be able to figure out whatever you need to. Sławomir Biały  (talk) 00:10, 16 October 2012 (UTC)
 * So $$\int f^{-1}(x)dx=x f^{-1}(x)-g(f(x))+C$$; example: $$\int \arcsin{x} \, dx = x \arcsin{x} + \cos{(\arcsin{x})} + C = x \arcsin{x} + \sqrt{1 - \sin^2{(\arcsin{x})}} + C = x \arcsin{x} + \sqrt{1 - x^2} + C$$ Ssscienccce (talk) 06:33, 17 October 2012 (UTC)


 * Thanks. Am I correct in assuming there's a typo in your first equation, Ssscienccce -- f(x) should be f-1(x)? Duoduoduo (talk) 20:56, 17 October 2012 (UTC)