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What's the integral of $f(x)=(1+x^2)^{1/2}$? I tried making $x=\sin(t)$ and doing integration by substitution but I don't think I arrived to the correct answer. Does this perhaps require the use of hyperbolic functions?

Shobhit
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2 Answers2

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The choice for $x = \sin t$ would be appropriate if you had $\sqrt{1 - x^2}$.

But when you have an integrand containing $a^2 + x^2$, like here, where you have $\sqrt{ 1 + x^2}$, try substituting $x = a\tan t$.

In this case, that would be substituting $x = \tan t$. Then $dx = \sec^2 t\,dt$.

See Wikipedia for a quick refresher on the three most common trigonometric substitutions.

amWhy
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So how about $x = \tan(t)$:

$$ \int \sqrt{1 + \tan^2(t)} \sec^2(t)\; dt = \int \lvert\sec(t)\rvert\sec^2(t)\;dt. $$

Thomas
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  • I got to this point, but how do you solve the integral when it is |sect|(sect)^2 – Vladimir Nabokov Oct 13 '13 at 13:10
  • @LeonLeibovici: Assuming that $\sec(x) \geq 0$ you have the integral of $\sec^3(x)$. For this you can see: http://math.stackexchange.com/questions/154900/indefinite-integral-of-secant-cubed – Thomas Oct 13 '13 at 14:32