How to Integrate sqrt(x^2+1) Using the u-Substitution Method?

[subwaybusker]

Homework Statement

integral: sqrt(x^2+1)

Homework Equations

The Attempt at a Solution

I tried using x=tan[y], dx=sec^2[y]dy

Integral:
sqrt(tan^2[y]+1)sec^2[y]dy
=sec^3[y]dy
=(tan^2[y]+1)(sec[y])dy
=(tan^2[y]sec[y]+sec[y])dy

and then i was stuck.

 

[rock.freak667]

From

$$\int sec^3 y dy$$

Write it as

$$\int (secy) (sec^2y dy)$$

and then use integration by parts.

 

[Dick]

I would write sec(y)^3 as dy/cos(y)^3=cos(y)dy/cos(y)^4=cos(y)*dy/(1-sin(y)^2)^2. Putting u=sin(y) that becomes du/(1-u^2)^2. Now use partial fractions.

 

[Gib Z]

subwaybusker, Do you know about hyperbolic trig functions? I would let $x=\sinh u$.

 

[psykatic]

Well, my way of approaching this integral would be,
$$\int \sqrt{x^2+1} . 1$$

So, now the above integral can be solved by using integral by parts, try doing it!

P.S: Here,$$u=\sqrt{x^2+1}$$ and v=1.

 

[Dick]

Well, my way of approaching this integral would be,
$$\int \sqrt{x^2+1} . 1$$

So, now the above integral can be solved by using integral by parts, try doing it!

P.S: Here,$$u=\sqrt{x^2+1}$$ and v=1.

Ok, so after an integration by parts, now you have to integrate x^2/sqrt(x^2+1). How is that a simplification? I think you are doing the parts integration wrong. You need to have u=sqrt(x^2+1) and v=x.

 

[psykatic]

Nope, once you have,
$$\int \sqrt{x^2+1}. 1~ =~ x\sqrt{x^2+1}~-~\int \frac{x^2}{\sqrt{x^2+1}}$$

Add and subtract 1 in the numerator, that'll help you, I'd solve it further if you want!

 

[Gib Z]

Well, putting back in our differentails, which is a bad idea to forget, and evaluting the trivial integral, our integral I through your method is given by

$$I = \frac{(x^2+1)^{\frac{3}{2}}}{3} - I + \int \frac{1}{\sqrt{x^2+1}}} dx$$

So Now you have to integrate that last integral, which takes the same substitution as suggested earlier. This method is not the shortest way.

 

[psykatic]

Well, Gibz thanks for making me realize my mistake! There shouldn't have been any integral before $$x.\sqrt{x^2+1}$$, my bad :sad:

Thats already integrated, going according to the rule it'd be:
$$\int u.v=~u\int v -\int \frac{\delta u}{\delta x} \int v$$

so, considering,$$u=\sqrt{x^2+1}$$ and v=1,

it'd be
$$x.\sqrt{x^2+1} -\int\frac{1}{2\sqrt(x^2+1)}.2x.x$$
where, 2 gets canceled in the numerator and the denominator, and the sum continues..