The integral of the function \((\frac{1-x}{1+x})^{\frac{1}{2}}\) can be approached using trigonometric substitution rather than partial fractions, as the latter is not applicable due to the nature of the function. The discussion highlights that multiplying by \(\frac{\sqrt{1-x}}{\sqrt{1-x}}\) transforms the integral into a more manageable form, specifically \(\int \frac{1-x}{(1-x^{2})^{1/2}} dx\). This leads to known integrals such as \(\int \frac{1}{\sqrt{1 -x^2}}\, dx\) and \(\int \frac{x}{\sqrt{1 -x^2}} \, dx\), which can be solved using trigonometric identities.
PREREQUISITESStudents and educators in calculus, particularly those focusing on integration techniques, as well as anyone seeking to deepen their understanding of trigonometric integrals and rational function manipulation.
RUber said:What if you multiplied by 1,
i.e.
## 1= \frac{\sqrt{1-x}}{\sqrt{1-x}}##
Then you might have something that looks like a known integral using trig functions.
Partial fractions isn't appropriate here, either. Assuming you could break up ##(\frac{1-x}{1-x^2})^{1/2}## into ##(\frac{A}{1-x} + \frac B {1 + x})^{1/2}##, you still have the sum of the two fractions being raised to the 1/2 power.binbagsss] How do I go about integrating ##(\frac{1-x}{1+x})^{\frac{1}{2}} ##? [h2]The Attempt at a Solution[/h2] im not really sure. could integrate by partial fractions if it was to the power of ##1## said:oh thanks, so by doing that I get ##\int \frac{1-x}{(1-x^{2})^{1/2}} dx ##, which I can partial fraction and then integrate