Integrating \sqrt{1+u^2}: Help Needed

[flash]

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

 

[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.

 

[flash]

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

 

[flash]

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

 

[arildno]

Sure you can.
Yields the same result.

 

[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.

 

[flash]

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

 

[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.

 

[Shadowz]

Does anyone know any tricks or how to integrate this function from negative inf to positive inf?
$$\int\\sinc( 2*pi*W*(t - k/(2*W)) ) * sinc( 2*pi*W*(t - j/(2*W)) )dt$$

Thanks.

 

[Gib Z]

The normalized or unnormalized sinc function?