integral2

**alex83** · April 24th 2009, 01:43 PM

integral

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

any help pls

**redsoxfan325** · April 24th 2009, 01:49 PM

Originally Posted by alex83

integral
$<br /> \int \frac{du}{\sqrt{1+u^2}}$

any help pls

You can find this in a table of integrals. It's $arcsinh(u)$ . (It's the "inverse hyperbolic sine of u".)

**Jameson** · April 24th 2009, 01:53 PM

Originally Posted by alex83

integral

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

any help pls

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 $\sqrt{1+u^2}$ . Let theta be the angle such that $\tan(\theta)=u$ . You will also notice that $\cos(\theta)=\frac{1}{\sqrt{1+u^2}}$ , which makes an easy substitution. Make sure to calculate $d\theta$ in terms of u, then make the full substitution, integrate, then back substitute.

**Krizalid** · April 24th 2009, 05:00 PM

a classic problem, we can also put $v=u+\sqrt{1+u^2}.$

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