# Book on multivariable calculus

1. Dec 23, 2004

### brendan_foo

I am reading a book on multivariable calculus for my course and I have tried the question :

$$\int_{0}^{\frac{\pi}{4}} dx \int_{0}^{Sec(x)} y^3 dy$$

apparently the answer is 1/3..I have TRIED to get this answer... yet i cannot yield 1/3....help!!!

(anyone from the UK...I have a good A Level in maths {up to P3 for Pure and M3 for mechanics, grade A...i know iterated integrals but this one stumps me}...arse

Last edited: Dec 23, 2004
2. Dec 23, 2004

### eddo

I worked it out and got 1/3. Show your work so that I can show you where you've gone wrong.

3. Dec 23, 2004

### quasar987

Are you using the formula

$$\int sec^n u = \frac{1}{n-1}tgu \ sec^{n-2}u+\frac{n-2}{n-1}\int sec^{n-2}udu$$

?

Using this on $sec^{4}x$, I get 1/3 too.

Last edited: Dec 23, 2004
4. Dec 23, 2004

### learningphysics

You can do the problem with integration by parts.

5. Dec 24, 2004

### brendan_foo

I have never seen that formula before ever :)...

naturally i established the first integral, and ended up with

$$\frac{1}{4}\int_{0}^{\frac{\pi}{4}}sec^4 (x) dx$$
however beyond this i begin to lame up... doing it by parts usually ends up in misery and i usually end up going around in circles.

$$\frac{1}{4}\int_{0}^{\frac{\pi}{4}}(sec^2 (x))(sec^2 (x)) dx$$

Would this suffice in the first instance, and take it from parts there on in..

BK Flamer

Last edited: Dec 24, 2004
6. Dec 24, 2004

### learningphysics

Yes. Set $$u=sec^2(x)$$ and $$dv=sec^2(x)dx$$

You only have to do integration by parts this one time... continue, and tell us where you get stuck...

7. Dec 24, 2004

### t!m

If sec is raised to an even power, you can simply do it using substitution and identities. That's how I did it, and also got 1/3. You're second step is correct, now let u=tan(x). Anything?

8. Dec 24, 2004

### brendan_foo

Doing it as we speak....hang on guys... :D

9. Dec 24, 2004

### brendan_foo

ok so i've said :

$$u = \sec^2(x) \quad \quad \frac{du}{dx} = 2\sec^2 (x) \tan(x) \quad \quad \frac {dv}{dx} = sec^2 (x) \quad \quad v = \tan (x)$$

So therefore, by parts, the integral {excluding constants and what not for simplicities sake}

$$\int (sec^2 (x))(sec^2 (x)) dx = sec^2 (x)tan (x) - 2\int sec^2 (x) (tan^2 (x)) dx$$

Or have I made a right boob-up?

Last edited: Dec 24, 2004
10. Dec 24, 2004

### t!m

I believe you meant

$$u = \sec^2(x)$$

But you derived it correctly, so I'll assume that was a typo. Otherwise, your setup looks right to me. I'm gonna go ahead and post how I would have done it without integrating by parts, since parts seems to be the preferred method right now.

Last edited: Dec 24, 2004
11. Dec 24, 2004

### learningphysics

Everything is correct. Just integrate $$sec^2(x)tan^2(x)dx$$. Have a look at it for a while... hint: use the fact that $$sec^2(x)$$ is the derivative of $$tan(x)$$

12. Dec 24, 2004

### brendan_foo

So i try to evaluate that other integral, by parts too... the :

$$\int sec^2 (x) (tan^2 (x)) dx$$

if someone would give me a guide as to how many times im going to need to do this repeated parts thingy then I wont be so scared ;) :rofl:

$$u = \tan^2(x) \quad \quad\frac{du}{dx} = 2\tan (x) \sec^2(x) \quad \quad\frac {dv}{dx} = sec^2 (x) \quad \quad v = \tan (x)$$

Ok so far?!?!...i still get the feeling that its not going to terminate anytime soon

Last edited: Dec 24, 2004
13. Dec 24, 2004

### learningphysics

Not integration by parts here: I'll do a variable substitution here: let $$y=tan(x)$$ so $$dy=sec^2(x)dx$$ so, I can rewrite your integral as:

$$\int y^2dy$$

which becomes

$$(1/3)y^3=(1/3)tan^3(x)$$

14. Dec 24, 2004

### brendan_foo

Oooh yeah!! :D ha god dammit im a fool... Could someone show it to me by parts too, cos it was that method that was murdering me.

:tongue:

15. Dec 24, 2004

### t!m

Without integrating by parts, using only a trig identity and variable substitution:

$$\frac{1}{4}\int_{0}^{\frac{\pi}{4}}(sec^2 (x))(sec^2 (x)) dx =\frac{1}{4}\int_{0}^{\frac{\pi}{4}}(tan^2 (x)+1)(sec^2 (x)) dx$$

Letting $$u=tan(x); du=sec^2(x)dx$$

We have
$$\frac{1}{4}\int(u^2 +1)du =\frac{1}{4}(\frac{1}{3}u^3 +u| =(\frac{1}{12}tan^3 (x)+\frac{1}{4}tan(x)|_{0}^{\frac{\pi}{4}} =\frac{1}{3}$$

Seems easier to me.

16. Dec 24, 2004

### brendan_foo

Most definatly...

HOWEVER :D...

due to my sloppy notation and consumption of beer, i was missing out something that was so so simple.

After careful considering, i arrived at the integral :

$$\frac{1}{4}(sec^2 (x) tan (x) - \frac{2}{3} tan^3 (x))$$

And when the limits are put in I do believe it yields 1/3...

I am therefore a boob!!! :rofl:

17. Dec 24, 2004

### learningphysics

You're right! I didn't see the identity.

Last edited: Dec 24, 2004
18. Dec 24, 2004

### learningphysics

That's alright. I'm a boob too! Hopefully us boobs can still find the right answer going about it the hard way!

19. Dec 24, 2004

### brendan_foo

Thanks all who gave advice...much appreciate

Merry christmas all...

:rofl: