# Thread: Help with an integral?

1. ## Help with an integral?

Hi, Sorry but, would someone of you so kindly help me with a particular integral that I have problems with? I would deeply appreciate it.

Now here it goes.

Trigonometrical substitution.
S='integral of.'

S(4+x^2)^.5dx

that means, integral of the square root of 4 plus x squared

This particular case is of the form of S(a^2+x^2)^.5

thus we use

x=atan( t)

x=2tan(t )
dx= 2sec^2(t )dt

Thus

S((4+4tan^2(t ))^.5)*2sec^2 (t )dt

moving some numbers... (1+tan^2 (t )=sec^2( t))
4 S ((sec^2( t))^.5)*sec^2( t)
then I simplified to

4*( S sec^3( t)dt)

then I got stuck. Help please? Am I doing something wrong?  Reply With Quote

2. ## Re: Help with an integral?

did you try to take the antiderivaive in steps?

leaving dx off the table, I got this

S(4+x^2)0.5

=> 2/3(4+x^2)^(3/2) . 4x+?X^3

so basically, I took the derivative of the outside, then the inside. and yes you can simplify this, but I'm too lazy to finish. you can substitute x for a trigonometric, sin or cos I dunno or care.

the ? is because I'm not sure what to put there-not terribly good at calculus  Reply With Quote

3. ## Re: Help with an integral?

I think you're correct so far. The integral of sec cubed is tricky but it's done on wikipedia. Then you'll need to express t in terms of x to get the answer.  Reply With Quote

4. ## Re: Help with an integral?

I'm proud to say that I am finish with Calculus. I aced the classes in both high school and university. I burned the textbooks afterwards.   Reply With Quote

5. ## Re: Help with an integral?

Have you tried using Wolfram Alpha?  Reply With Quote

6. ## Re: Help with an integral?

Thanks for your help guys. I already sorted it out. It's not that hard now, though it is a little more elaborated.

4*((sec(t))^3 dt= 4*((sec(t))^2*sec(t) dt) Then you just integrate 'by parts' or something like that, I dont remember the specific name but it's Sudv=uv-Svdu. Then it's just a matter of playing with the coeficients in both sides of the equation.   Reply With Quote

7. ## Re: Help with an integral?

Could you please post the answer? I'm breaking my head here trying to solve the integral. And to think that I just finished revising basic Calculus for tomorrow's exam...

Maion  Reply With Quote

8. ## Re: Help with an integral?

Ok this is sort of spam but...

...I am SOOOOO glad that none of this makes sense to me anymore. It really amazes me how out of touch the stuff we had to do at school is with real life...

...2 Uni degrees and I almost never had to use any of the advanced maths I did in school.

Maion I hope you did great with your exam. 3 lykeiou?  Reply With Quote

9. ## AW: Help with an integral?

Dear Jesus. Do you know what formulae are for me? They are tiny abstract paintings in black and white, printed on bad paper.  Reply With Quote

10. ## Re: Help with an integral? Originally Posted by rasoforos Ok this is sort of spam but...

...I am SOOOOO glad that none of this makes sense to me anymore. It really amazes me how out of touch the stuff we had to do at school is with real life...

...2 Uni degrees and I almost never had to use any of the advanced maths I did in school.

Maion I hope you did great with your exam. 3 lykeiou?
Indeed, I agree that our educational system is deeply rotten. A shame, for we have a great potential as a nation. As for the Mathematics themselves, things like Rolle's, Bolzano's Theorems and other pointless theorems we learned (by heart) for estimating the possible number of roots for a given function springs to mind. Our mistake, as see it, is that we do way too theoretical stuff. We have to teach children to be practical early on. I was surprised to see how important Calculus is for solving a multitude of everyday problems, not to mention the power of several of its computational applications (which are plentiful in my textbooks).

And I did good in my exams last year, thank you. I'm now officially a student! Anyway, as for the integral itself I decided to call upon some "higher" help. Mathematica gave me the result: Where [sin(x/2)]^(-1) is of course the inverse sine function (or arcsin). A simple differentiation of the result gives us the integrated function, which basically stands as proof of our result.

I'm having my next Calculus exams the day after tomorrow, which basically has to do with multi-variable functions and multiple integrals. But I will surely try and find an ample solution to this problem once I find the time.

Maion  Reply With Quote

11. ## Re: Help with an integral? Originally Posted by rasoforos Ok this is sort of spam but...

...I am SOOOOO glad that none of this makes sense to me anymore. It really amazes me how out of touch the stuff we had to do at school is with real life...

...2 Uni degrees and I almost never had to use any of the advanced maths I did in school.

Maion I hope you did great with your exam. 3 lykeiou?
Then you have not been sitting in the class room finding immediate use of what the teacher just taught. I couldn't figure out to make my spheres go in perfect sircles in the simple Python script i was making. Then this wonderful trigonometry chapter came; teacher explaining that the cosine and sine of the degree we are creating in the unit circle will create the coordinates of a circle. What a wonderfully simple way to draw a sircle. I also, thanks to purely theoretical mathematics, realised how to draw 3D objects on a computer using mathematical formulae; letting them be made out of x amount of spheres, each who's radius equals y of the formula (using integrals to calculate volume).

All it takes is a creative mind to find use of it. :-P  Reply With Quote

12. ## Re: Help with an integral?

I think calculus should be taught within its background first and how it solves some classical problems.

I would start with simplified calculus to show acceleration, force and energy equations for physics. To show its strengths right off.

I would then use it to resolve problems that arithmetic cannot solves. Classic problems such as Zeno's paradox of Hercules and the Tortise.

And then build up from there.

But I would also teach the beauty and wonder in maths itself.  Reply With Quote

13. ## Re: Help with an integral?

Indeed, more emphasis should be put on the historical evolution of Calculus, which is in truth a long, hard and painful road of many failures. Archimedes made the first steps of Calculus by using the exhaust method for deriving certains volumes of solids, plus appriximating the number Pi. Later on his works were lost, re-found and taken some steps forward by Arab and Persian Mathematicians. But, in the end, it would be the (admittingly weird at first contact) concept of the infinite that caused the revolution in Infinitesimal Calculus as we know it.

Needless to say, I was thrilled to use Calculus to derive volumes, centers of mass, moments of interia and other nifty stuff used in our everyday lives as well as Physics which is my subject of study after all. A truly wonderful tool for being able to calculate many things, or approximate changes and behaviour of various phenomena taking known symmetries in mind.

Maion  Reply With Quote

14. ## Re: Help with an integral?

The integral of a circle's circumference is its area, and the volume of a sphere is the integral of the circles area.

Very neat way of showing how integrals evaluate real life objects.  Reply With Quote

15. ## Re: Help with an integral?

Indeed, plus infinitesimally thin disks and cylindrical shells are also nifty tricks of calculating volumes of certain solids that revolve around an axis. Needless to say, I was more than thrilled to find out how to derive volumes, surface areas, moments of interia etc. of many known solids myself.

Not to mention the importance and wide usage of polar (in 2D) coordinates, as well as spherical and cylindrical (in 3D). Really, some problems are nigh-impossible to solve without switching from Cartesian coordinates to one of the aforementioned.

Maion  Reply With Quote

16. ## Re: Help with an integral?

Sorry about how late this is, but I've been gone for a loooooooong time. Hi to everyone who remembers me! (if anyone does - I'm Tiberius)

This is quite simple if you've learned hyperbolics:

You substitute x for something else:
x = 2sinhy

Then you get:
= 2S (1+sinh^y) dy 2coshy
= S 2(coshy)^2 dy
= S cosh2y + 1 dy
= 0.5sinh2y + y (+c)

It's just using these hyperbolic identities:
cosh2y = (coshx)^2 + (sinhx)^2
1 = (coshx)^2 - (sinhx)^2  Reply With Quote

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