# Integration by Parts definite integral

## Homework Statement

The definite integral of from 0 to 1 of ∫ (r3)dr/sqrt(4+r2)

## Homework Equations

∫udv = uv - ∫vdu

∫du/sqrt(a2 - u2) = arcsin(u/a) + C

∫du/(asqrt(a2 - u2)) = (1/a)arcsec(u/a) + C

## The Attempt at a Solution

because I thought it easier to get it's derivative, rather than integral by making it dv.

du = -r(4+r2)-3/2dr

dv = r3dr
v = (1/4)r4

I've set up the equation as
= uv -∫vdu
= r4/(4(4+r2)1/2) + ∫r3dr/(4+r2)3/2

This seems somewhat similar to the the second equation I put in "Relevant Equations" (where I could put a two in the numerator and multiply the integral by (1/2), but both top and bottom are to the power of 3 (which I don't know how to get rid of. AND the bottom is not a2 - u2 but a2 + u2.

If somebody could please help me understand this I would be greatly appreciative. I'd like to be able to master this type of integration, but I seem to have hit a wall.

Dick
Homework Helper
It's a LOT easier if you just work it through with the u substitution u=4+r^2.

hunt_mat
Homework Helper
What about letting u=x^3 and v=1/sqrt(4+r^2): to obtain:
$$\int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}dr=\left[ r^{3}\sinh^{-1}(r/2)\right]_{0}^{1}-3\int_{0}^{1}r^{2}\sinh^{-1}(r/2)dr$$

What about letting u=x^3 and v=1/sqrt(4+r^2): to obtain:
$$\int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}dr=\left[ r^{3}\sinh^{-1}(r/2)\right]_{0}^{1}-3\int_{0}^{1}r^{2}\sinh^{-1}(r/2)dr$$
but according to integration by parts, 1/sqrt(4+r^2) must be dv not v, if you make u = x^3. Also I did not follow how you got the sinh^{-1}(r/2) if the denominator is (a2 + u2)1/2 not (a2 - u2)1/2

I see what you did for the most part, except that last issue I mentioned. Thnx.

hunt_mat
Homework Helper
Okay, look at the integral:
$$\int\frac{dx}{\sqrt{a^{2}+x^{2}}}$$
Use the substitution $$x=a\sinh u$$ and the integral reduces down to:
$$\int du=u=\sinh^{-1}\left(\frac{x}{a}\right)$$
It's a standard integral

HallsofIvy
Homework Helper
but according to integration by parts, 1/sqrt(4+r^2) must be dv not v
That was what hunt_mat meant.

, if you make u = x^3. Also I did not follow how you got the sinh^{-1}(r/2) if the denominator is (a2 + u2)1/2 not (a2 - u2)1/2
That is exactly the point! If it were $\sqrt{a^2- u^2}$, then the integral would be $sin^{-1}(x)$. But because it is "+" rather than "-" it is the hyperbolic rather than regular sine.

I see what you did for the most part, except that last issue I mentioned. Thnx.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

The definite integral of from 0 to 1 of ∫ (r3)dr/sqrt(4+r2)
...

## The Attempt at a Solution

because I thought it easier to get it's derivative, rather than integral by making it dv.

du = -r(4+r2)-3/2dr

dv = r3dr
v = (1/4)r4

I've set up the equation as
= uv -∫vdu
= r4/(4(4+r2)1/2) + ∫r3dr/(4+r2)3/2

This seems somewhat similar to the the second equation I put in "Relevant Equations" (where I could put a two in the numerator and multiply the integral by (1/2), but both top and bottom are to the power of 3 (which I don't know how to get rid of. AND the bottom is not a2 - u2 but a2 + u2.

If somebody could please help me understand this I would be greatly appreciative. I'd like to be able to master this type of integration, but I seem to have hit a wall.
The integral that resulted when you did integration by parts,

$$\int_{0}^{1}\frac{r^{3}}{(4+r^{2})^{3/2}}dr\,,\$$ is more difficult to integrate than the integral you started with:

$$\int \frac{r^{3}}{(4+r^{2})^{1/2}}dr\,.$$

This suggests reversing the roles of u & v to the extent that you can do that. Integrating $$(4+r^{2})^{-1/2}$$ will result in something with a factor of $$(4+r^{2})^{1/2}$$. In fact, although this is a little unusual, you can pick what you want v to be, then differentiate that to see if the dv which results is usable.

Let $$\textstyle v=(4+r^{2})^{1/2}$$, then $$\textstyle dv=r(4+r^{2})^{-1/2}\, dr\,.$$ This will work out just fine!

Therefore, let $$\textstyle dv=r(4+r^{2})^{-1/2}\,dr\ \ \to\ \ v=(4+r^{2})^{1/2}\,.$$

This leaves $$\textstyle u=r^2\ \ \to\ \ du=2r\,dr,.$$

$$\int u\ dv=\int (r^2)\left(r(4+r^{2})^{-1/2}\right)\,dr$$ $$\text{, and } \int v\ du=2\int (r)\left((4+r^{2})^{1/2}\right)\,dr$$

The integral $$\int r\,(4+r^{2})^{1/2}\,dr$$ is easily evaluated with a simple substitution.