# Thread: Calculus

1. just until yesterday i thought i was alright at calculus until i saw this very fundamental calculus question which i couldn't solve.

can anyone help me?

thank you.

2.

3.

4. I think you have to change the equation to be expressed in terms of theta....
(runs to get her calc book)
I just got a C in my calc III class, I was disappointed. My teacher sucked.

5. taking a quick glance at it, id say to use Partial Fractions then integrate as two separate integrals

1/(1+cos(x)) and 1/(1+sin(x))

using MAPLE, integrating produces

tan(1/2(x)) - 2/[tan(1/2(x)) + 1] evaluated from 0 to pi

6. Originally Posted by Heinsbergrelatz
just until yesterday i thought i was alright at calculus until i saw this very fundamental calculus question which i couldn't solve.

can anyone help me?

thank you.

=
=
=
=
=1

7. Originally Posted by DrRocket

=
Doesn't that denominator equal 1 + cos(x) + sin(x) + cos(x)sin(x)?

8. it does, I have a feeling Dr. Rocket intended to make that mistake to see if we were paying attention.

let then;

so;

And I believe this is much easier to solve, provided I made no mistakes. And it is;

9. Originally Posted by pbandjay
Originally Posted by DrRocket

=
Doesn't that denominator equal 1 + cos(x) + sin(x) + cos(x)sin(x)?
Yep, I must have been tired when I looked at the problem.

But the integral after the mistake is more interesting.

10. Originally Posted by Arcane_Mathematician

Could someone explain this step?

I'm not very good at this trig stuff
OK I found sin 2 x = 2 sin x cos x which might help.

I would go

8( 2/sin^2@ - 1/sin2@ - 2root(2)/cos^2@sin@)

OR 8( 2/sin^2@ - 1/2sin@cos@ - 2root(2)/cos^2@sin@)

Sorry I can't use the 'tex' thing properly!!

Trying now!!

Note the @ are theta (now fixed!!)

Actually when I first though it I though this is like integrating 1/u to log(u)

Could I do that?

When I say 'I' I mean could anyone (who is better at maths than me) do that?

11. Originally Posted by smokey
Originally Posted by Arcane_Mathematician

Could someone explain this step?
Originally Posted by Arcane_Mathematician
it does, I have a feeling Dr. Rocket intended to make that mistake to see if we were paying attention.

let then;

the first half of my post was the explanation of that step

12. Originally Posted by Arcane_Mathematician

And I believe this is much easier to solve, provided I made no mistakes. And it is;

I meant this bit, seems I may have quoted the wrong line.

For the second line I would get

13. Thank you smokey. I made an arithmetic error earlier that I didn't catch. Working it all out here, I believe I was able to see it.

14. Sorry for my late appreciation to all the members who helped me with this question, it was cause i was going to solve mine and compare the workings.

anyway i got to a similar approach and got a similar answer, but not exactly same, but i will still work on that

anyway thanks alot.

15. post your work, I'm not 100% sure I did it right, and your answer may be more solid.

16. its really a informative post...
thanks for your information ...........

17. then you just apply the limits. does it look alright?

18. how did you get to that starting point?

19. how did you get to that starting point?
i just did simplified it using partial fractions from the original integral.

20. Originally Posted by Heinsbergrelatz
how did you get to that starting point?
i just did simplified it using partial fractions from the original integral.
I can't figure out how you used partial fractions for this one. It isn't registering for me, could you post all of the steps you used to arrive at this conclusion?

21. Hey arcane, here is my working alittle modified and easier actually, though the post is a little late, just see if you agree with it.

okay this little nasty integral needed some decent amount of trigonometric to "t-substitution method"

okay because now we have inserted:

its pretty tedious to try and make "t" the subject of the formula and plug it in to our integral right? so how about this?

since the bounds were

so our new Up. and Lower bounds are:

i would appreciate any corrections to my mistakes.

22. +12=13

23. Originally Posted by ismaeljohnson11
+12=13
No.

Proof:

QED

Rather, +12 is all values that are greater than 12. If we consider the natural numbers than it would be the set A, for all A = {13, 14, 15,...}. On the other hand, if we are considering real numbers, than it would be x, for all x greater than 12, that is to say decimals other than zero are allowed (e.g., 12.1, 12.01, 12.001, etc...).

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