Why Can't I Solve This Multivariable Calculus Integral?

[brendan_foo]

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

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

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

 

[eddo]

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

 

[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.

 

[learningphysics]

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.

You can do the problem with integration by parts.

 

[brendan_foo]

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

naturally i established the first integral, and ended up with

$$<br /> \frac{1}{4}\int_{0}^{\frac{\pi}{4}}sec^4 (x) dx$$
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.

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

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

BK Flamer

 

[learningphysics]

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

naturally i established the first integral, and ended up with

$$<br /> \frac{1}{4}\int_{0}^{\frac{\pi}{4}}sec^4 (x) dx$$
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.

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

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

BK Flamer

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...

 

[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?

 

[brendan_foo]

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

 

[brendan_foo]

ok so I've said :

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

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

$$<br /> \int (sec^2 (x))(sec^2 (x)) dx = sec^2 (x)tan (x) - 2\int sec^2 (x) (tan^2 (x)) dx<br />$$

Or have I made a right boob-up?

 

[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 going to go ahead and post how I would have done it without integrating by parts, since parts seems to be the preferred method right now.

 

[learningphysics]

ok so I've said :

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

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

$$<br /> \int (sec^2 (x))(sec^2 (x)) dx = sec^2 (x)tan (x) - 2\int sec^2 (x) (tan^2 (x)) dx<br />$$

Or have I made a right boob-up?

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 $$tan(x)$$

 

[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 I am going to need to do this repeated parts thingy then I won't be so scared ;)

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

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

 

[learningphysics]

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 I am going to need to do this repeated parts thingy then I won't be so scared ;)

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

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

 

[brendan_foo]

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

 

[t!m]

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

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

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

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

Seems easier to me.

 

[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 :


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

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

I am therefore a boob!

 

[learningphysics]

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

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

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

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

Seems easier to me.

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

 

[learningphysics]

I am therefore a boob!

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

 

[brendan_foo]

Thanks all who gave advice...much appreciate

Merry christmas all...