# Evaluating Integral

1. Feb 24, 2007

### clipzfan611

1. The problem statement, all variables and given/known data
Integral: Upper limit is pie/3, lower limit is 0. secxtanx(square root of 1+secx)dx

Sorry if this is confusing, I dont know how to use the real math symbols.

3. The attempt at a solution
I realized it is U-substitution and not the FTC.
u=1+secx dx=sextanx du
when x=0, u=2 & x=pie/3, u=3.
I dont really know what to do after that, I'm a total noob at u-sub. Please help.

2. Feb 24, 2007

### Gib Z

$$\int_0^{\pi/3} \sec x \tan x \sqrt{1+\sec x} dx$$.

Correct, let u = 1+sec x.

sec x tan x dx=du

So now its

$$\int \sqrt{u} du = \frac{2}{3} \cdot u^{\frac{3}{2}}$$

Replace back in u=1+sec x
The integral is $$\frac{2}{3} \cdot (1+\sec x)^{\frac{3}{2}}$$

Now just put back in the bounds of integration, as original, and evaluate. If only have to change the bounds when you want to keep the u at the end, but sometimes just makes it more complicating.

Edit: Im not bothering to see if you change your bounds correctly, just assuming they are, then you can sub in u=3 minus the u=2 of $$\frac{2}{3} \cdot u^{\frac{3}{2}}$$

Last edited: Feb 24, 2007
3. Feb 24, 2007

### clipzfan611

Thank you.

I got:
Does that seem correct?

4. Feb 24, 2007

### Gib Z

Which way did you get it? With the u's, or did you change back to the trig stuff?

with the u's it should be

$$\frac{2}{3}3^{\frac{3}{2}} - \frac{2}{3}2^{\frac{3}{2}}$$
In the first part, do you notice, ignore the factor of 2 right now, 3^(1.5) /3, that simplifies to root 3. doing simple things like that, you should have gotten $$2\sqrt{3} - \frac{4}{3}\sqrt{2}$$.

5. Feb 24, 2007

### clipzfan611

With the u's. Are you sure that you multiply 2/3 to both, or the difference of the two?

6. Feb 24, 2007

### Gib Z

It should make no difference as the distributive property holds:
In other words
$$\frac{2}{3}(a-b) = \frac{2}{3}a - \frac{2}{3}b$$

Last edited: Feb 24, 2007
7. Feb 24, 2007

### clipzfan611

I figured because of the multiplication dot that you had to do what's in paranthesis first, but it wasnt subtractable so sorry for the confusion.

Thanks for your help, you were really patient.

8. Feb 24, 2007

### Gib Z

Your very welcome. I have all day, so no reason for me to be impatient :)

By the way, Welcome to Physicsforums !