# Integration By Parts and Substitution

1. Sep 18, 2007

### Lanza52

[SOLVED] Integration By Parts and Substitution

Short background; Took Calc 1 my senior year in highschool. Got As all 4 quarters and found it quite easy. Freshman year comes around and I sign up for Calc 2. Turns out the only teacher teaching Calculus 2 for my fall and spring semester is a horrible teacher who doesn't care about his students. So I decided to wait until my Sophomore year to take Calc 2.

So fast forward; I am picking up on all the new techniques and what have you, but I keep forgetting all the old, little essential things because of that year off.

So heres the problem;

$$\int\pi^{\sqrt{x}}dx$$

Solve using integration by parts. At the bottom of the problem my professor gave a hint saying "You may like to do a substitution first."

And I'm stumped. I haven't used substitution at all in the past 15 months.

The only substitution I can think of is $${\sqrt{x}}=m$$, but that is of no use.

2. Sep 18, 2007

### morphism

Well,

$$\pi^{\sqrt{x}} = e^{\sqrt{x} \, \ln \pi}$$

And maybe the substitution u=\sqrt{x} from here?

3. Sep 18, 2007

### Lanza52

Played around with it for a while, didn't find anything. Not saying your wrong tho, I'm probably wrong.

4. Sep 18, 2007

### morphism

I decided to use the substitution $u=\ln(\pi)\sqrt{x}$ instead, but that doesn't really matter. This transforms the integral into

$$\frac{2}{(\ln{\pi})^2} \int u e^u du.$$

And from here it's an easy application of integration by parts.

5. Sep 18, 2007

### Lanza52

lol thanks for the help. But this is the part I don't remember how to do. Can't figure out how to get from
$$\int{\pi^\sqrt{x}}\ dx$$
to
$$\frac{2}{(\ln{\pi})^2} \int u e^u du.$$

6. Sep 18, 2007

### Dick

pi^(sqrt(x))=exp(ln(pi)*sqrt(x)). Then u=ln(pi)*sqrt(x), du=ln(pi)*dx/(2*sqrt(x)). Please check this and then try to continue.

7. Sep 18, 2007

### Lanza52

$$u=ln\pi\sqrt{x}$$
$$du=\frac{ln\pi dx}{2\sqrt{x}}$$

If I'm not wrong, that substitution would require;

$$\int\frac{ln\pi}{2\sqrt{x}} e^\sqrt{x}ln\pi$$

I can account for the $$\frac{ln\pi}{2}$$ but I don't remember how to get the $$\frac{1}{\sqrt{x}}$$ before the e in the integral so that you can apply the du.

8. Sep 18, 2007

### Dick

You are wrong. The integral is exp(u)*dx. Would you please solve for dx in terms of u and du so we can get on with this? Sorry to be snappy, it's getting late.

9. Sep 19, 2007

### Lanza52

Still drawing a blank.

And I've never seen the term exp(u). I was thinking of it as e^(u) but the more I play with the numbers the more I see I'm wrong.

10. Sep 19, 2007

### morphism

exp(u) is indeed just e^u.

We have:
du=ln(pi)*dx/(2*sqrt(x))

and

u=ln(pi)*sqrt(x).

It shouldn't be hard to write dx in terms of u and du.

11. Sep 20, 2007

### Lanza52

Okay, pretty sure I got it. Like I said, I was just forgetting simple little substitution rules.

Thanks.

12. Sep 20, 2007

### Lanza52

Okay, came up with a solution (I think)

Oh yea, the original problem was from 1 to 4. $$\int\pi^{\sqrt{x}}$$

u = m dv = e^m dm
du = dm v = e^m

me^m - $$\int$$ e^m dm

And from there subtract.

I came to 12.27177549 but this little online integral solver is telling me otherwise. (http://www.solvemymath.com/online_math_calculator/integrals/definite_integral_solver.php [Broken])

Last edited by a moderator: May 3, 2017
13. Sep 20, 2007

### Dick

Ok, so u*e^u-e^u is ok. But you've lost all of the multiplicative constants like ln(pi) and 2. Where did they go?

14. Sep 20, 2007

### Lanza52

Ahhhh. Thank you very much.