# Integration help

1. Aug 26, 2007

### flash

Hi
I am trying to integrate:
$$\sqrt{1+u^2}$$
It looks simple but it's causing me alot of problems. I've tried substitution, and by parts but can't get it. Thanks for any help!

2. Aug 26, 2007

### arildno

You can try $$u=Sinh(v)\equiv\frac{e^{v}-e^{-v}}{2}$$

You thereby get:
$$\frac{du}{dv}=Cosh(v)\equiv\frac{e^{v}+e^{-v}}{2}\to{du}=\Cosh(v)dv$$

Since you may verify that the hyperbolic functions satisfy the identity $Cosh^{2}(v)-Sinh^{2}(v)=1$, your integral is readily transformed to:
$$\int\sqrt{1+u^{2}}du=\int\sqrt{1+Sinh^{2}(v)}Cosh(v)dv=\int{Cosh^{2}(v)}dv$$ which is easily integrated.

3. Aug 26, 2007

### flash

Thanks. So I have to use hyperbolic functions to do it?

4. Aug 26, 2007

### flash

I just dug this up...do you think I could use it?
$$\int\sqrt{a^2+u^{2}}du=\frac{u}{2}\sqrt{a^2+u^2}+\frac{a^2}{2}ln |u + \sqrt{a^2+u^2}|$$

5. Aug 26, 2007

### arildno

Sure you can.
Yields the same result.

6. Aug 26, 2007

### arildno

To see that it is the same, let's compute the integral with hyperbolic substitution:

$$I=\int{C}osh^{2}(v)dv=Sinh(v)Cosh(v)-\int{S}inh^{2}(v)dv=Sinh(v)Cosh(v)-\int(Cosh^{2}(v)-1)dv=$$
$$Sinh(v)Cosh(v)+v-\int{C}osh^{2}(v)dv=Sinh(v)Cosh(v)+v-I$$

That is, we have the expression:
$$I=Sinh(v)Cosh(v)+v-I$$
yielding:
$$I=\frac{1}{2}Sinh(v)Cosh(v)+\frac{1}{2}v$$

Now, we have $$v=Sinh^{-1}(u)$$

Using the identity:
$$Cosh^{2}(y)-Sinh^{2}(y)=1$$
we get $$Cosh(Sinh^{-1}(u))=\sqrt{1+u^{2}}$$
since we have, of course $$Sinh(Sinh^{-1}(u))=u$$

Thus, our first term may be rewritten as:
$$\frac{1}{2}Sinh(v)Cosh(v)=\frac{u}{2}\sqrt{1+u^{2}}$$

Using the exponential representation of the the hyperbolic sine, you should have little trouble representing the inverse function of hypsine in terms of the natural logarithm.
That constitutes the second term in the formula you found.

Last edited: Aug 26, 2007
7. Aug 26, 2007

### flash

Awesome. Thanks heaps for your help! I should be right to do it now.

8. Aug 26, 2007

### Gib Z

Basically there are 2 ways of doing this integral: A normal trig substitution, u= tan x, or the hyperbolic trig one arildno suggested. The hyperbolic one is faster, but unless you are familiar with hyperbolic trig functions it is harder to spot when to use them as a substitution. They used a more general substitution for the formula you gave: u= a tan x.

9. Aug 28, 2007

$$\int\\sinc( 2*pi*W*(t - k/(2*W)) ) * sinc( 2*pi*W*(t - j/(2*W)) )dt$$