How Do You Integrate x^2 Over the Square Root of (x^2 + 3)?

[ParoXsitiC]

Homework Statement

\int \frac{x^2}{\sqrt{x^2+3}}

Homework Equations

sinh-1(u) = u' / (u^2 + 1)

The Attempt at a Solution

Make the x^2 + 3 look like x^2 + 1 by taking out a sqrt(3). Giving you

\int \frac{x^2}{\sqrt{3} \sqrt{\frac{x^2}{3}+1}}

Set the constant outside the integral.

\frac{1}{\sqrt{3}} \int \frac{x^2}{\sqrt{\frac{x^2}{3}+1}}

Now we find where u^2 = \frac{x^2}{3} , which is u = \frac{x}{\sqrt{3}}. Now we know the u of the sinh-1, we find u'

u' = \frac{1}{\sqrt{3}}So now we taken care of everything but x^2...

Where to go now?

 

[micromass]

So, now substite in the sinh. You'll see that it works out fine...

What do you get??

 

[ParoXsitiC]

I know that the sinh^-1(x/sqrt(3)) is in the answer, but there is still a multiplication between x^2 and the the sin so more needs to be done.

I can't just have

(1/3)x^3 * sinh^-1(x/sqrt(3)) + c

because the product rule states uv' + u'v, if it was as easy as u'v' then it would work.

 

[micromass]

I know that the sinh^-1(x/sqrt(3)) is in the answer, but there is still a multiplication between x^2 and the the sin so more needs to be done.

I can't just have

(1/3)x^3 * sinh^-1(x/sqrt(3)) + c

because the product rule states uv' + u'v, if it was as easy as u'v' then it would work.

You have the integral

\int{\frac{x^2}{\sqrt{x^2+1}}dx}

after your substitutions right?? So just substitute in x=sinh(u)

 

[HallsofIvy]

Micromass likes his hyperbolic functions substitutions! Perhaps just because i learned them first, I always think of trig substitutions first. Here, after taking out a \sqrt{3}, you have \sqrt{1+(x/\sqrt{3})^2} and since 1+ tan^2(\theta)= sec^2(\theta) I would let x= \sqrt{3}tan(\theta).

Of course, then, the x^2 in the numerator becomes 3tan^2(\theta)

 

[ParoXsitiC]

\sqrt{1+(x/\sqrt{3})^2}

Shouldnt that be 3 not sqrt(3)

I still don't get it. I need more of a step by step. coming to the solution wolfgram comes to:

I see how you decided what x is, but I don't see how they decided what u was.

 

[HallsofIvy]

\sqrt{1+(x/\sqrt{3})^2}

Shouldnt that be 3 not sqrt(3)

No, it shouldn't. Notice the parentheses. Both x and \sqrt{3} are squared.

I still don't get it. I need more of a step by step. coming to the solution wolfgram comes to:

I see how you decided what x is, but I don't see how they decided what u was.

The crucial point is that you have \sqrt{1+ a^2} and want to get rid of the square root. You should immediately think of the trig identity 1+ tan^2\theta= sec^2\theta.