Evaluate Integral: Upper Limit Pie/3, Lower 0 - U-Substitution

[clipzfan611]

Homework Statement

Integral: Upper limit is pie/3, lower limit is 0. secxtanx(square root of 1+secx)dx

Sorry if this is confusing, I don't know how to use the real math symbols.

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 don't really know what to do after that, I'm a total noob at u-sub. Please help.

 

[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: I am 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 $$\frac{2}{3} \cdot u^{\frac{3}{2}}$$

 

[clipzfan611]

Thank you.

I got:
2/3(3(radical 3)-2(radical 2)
Does that seem correct?

 

[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 $$2\sqrt{3} - \frac{4}{3}\sqrt{2}$$.

 

[clipzfan611]

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 $$2\sqrt{3} - \frac{4}{3}\sqrt{2}$$.

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

 

[Gib Z]

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

 

[clipzfan611]

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

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.

 

[Gib Z]

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

By the way, Welcome to Physicsforums !