integral

du/(sqrt(1+u^2) )

any help pls

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- Apr 24th 2009, 01:43 PM #1

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- Apr 24th 2009, 01:49 PM #2

- Apr 24th 2009, 01:53 PM #3

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As you'll see this is a standard trig integral, but it's better to do it by trig substitution.

Construct a right triangle with a vertical leg of u, a horizontal leg of 1 and by Pythagoras the hypotenuse is $\displaystyle \sqrt{1+u^2}$. Let theta be the angle such that $\displaystyle \tan(\theta)=u$. You will also notice that $\displaystyle \cos(\theta)=\frac{1}{\sqrt{1+u^2}}$, which makes an easy substitution. Make sure to calculate $\displaystyle d\theta$ in terms of u, then make the full substitution, integrate, then back substitute.

- Apr 24th 2009, 05:00 PM #4