Integration by Substitution

**Apex** · October 30th 2008, 02:59 PM

Hi

Integral, I = -4 * dx/(3+2x-x^2)^0.5
There is an integral sign between 4 and dx.

I = -4 * dx/((4-(x-1)^2)^0.5 <--- completing the square

Let x-1 = 2sin(#), where # is theta

The result is 4arcsin((x-1)/2) + c, where c is an arbitrary constant

Please explain how this works.

Thanks

**BoboStrategy** · October 30th 2008, 03:56 PM

$ -4 * \int{\frac{dx}{\sqrt{((4-(x-1)^2))}}} = -4 * \int{\frac{dx}{\sqrt{((2^2-(x-1)^2))}}} $

Then, since
$ \int{\frac{du}{\sqrt{a^2-u^2}}} = arcsin \frac{u}{a} + C, $

just let $u = x - 1$ and then we have

$ -4 * \int{\frac{dx}{\sqrt{((2^2-(x-1)^2))}}} = -4 * arcsin \frac{x-1}{2} + C $

**Apex** · October 30th 2008, 04:14 PM

Thanks!

**Apex** · October 30th 2008, 04:47 PM

The method in the notes is "let (x-1) = 2sin(theta)" for the above.

How does this work?
Can you show this using Pythagoras theorem?
Why is it arcsin rather than sin?

Thanks again

**BoboStrategy** · October 30th 2008, 05:01 PM

I don't see how letting

$ (x-1) = 2 * sin(\theta) $

is useful in this context.

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