Maths Gurus - Help a brother out?
Howdy,
So I've got these two Maths questions that I'm having a reeeallllyyyy hard time solving. I'm alright at Maths, but it's certainly not my best subject. So I was wondering whether you guys could lend a hand?
Here we go:
Question 1:
Question 2:
Any help most appreciated (I give cookies out for help ).
Those two look serious intimidating but really there's one trick for each one that makes it much easier:
for problem 1, try integration by parts (think about what u and v could be to get the right u*dv and v*du for this to work)
http://en.wikipedia.org/wiki/Integration_by_parts
for problem 2, you need the method of partial fractions in order to break apart the fraction into the sum of terms that are easy to integrate. Presumably you could also use long division of polynomials to figure it out, too, but who ever remembers how to do long division?
http://en.wikipedia.org/wiki/Partial_fraction
edit: actually this will probably be more helpful than the above link: http://en.wikipedia.org/wiki/Partial_fractions_in_integration
These hints should get you started - let us know if there's any more hints needed:)
for problem 1, try integration by parts (think about what u and v could be to get the right u*dv and v*du for this to work)
http://en.wikipedia.org/wiki/Integration_by_parts
for problem 2, you need the method of partial fractions in order to break apart the fraction into the sum of terms that are easy to integrate. Presumably you could also use long division of polynomials to figure it out, too, but who ever remembers how to do long division?
http://en.wikipedia.org/wiki/Partial_fraction
edit: actually this will probably be more helpful than the above link: http://en.wikipedia.org/wiki/Partial_fractions_in_integration
These hints should get you started - let us know if there's any more hints needed:)
I thought that'd be what you'd do, but I'm having trouble getting started (particularly the first one), because I have absolutely no idea what to do with the inverse Sine.
Oddly enough the first problem has nothing to do with inverse sin in particular. The only thing you need to know about inverse sin is its derivative: http://en.wikipedia.org/wiki/Inverse_tr ... _functions
So actually, think of sin^-1(pi*x) instead as some kind of generic function f(x). You can then figure out the derivative of f(x) and rewrite the integral in terms of f(x) and df(x)/dx, where it becomes clearer where to apply integral by parts.
DevilInPgh
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Joined: 23 Aug 2005
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Actually, the polynomial is in its completely reduced form. Instead, you have to use integration by parts. Once you factor the denominator, you will see how apparent u, du, v, and dv are. Additionally, this is a case in which you also have to integrate by substitution.
Also, the first one uses integration by substitution. Consult the trig tables for that.
Edit: You will also have to consult your trig/inverse-trig derivative tables. One key to integration in calculus is understanding patterns and thus being able to look something up very quickly. If I recall correctly (I took the AP Calc AB about 9 years ago, so I might be a little foggy on the exam experience), I think you're given a table with trig derivatives (but again, I don't know how it works in Australia).
Edit 2: Just a warning: the second one is going to take A LOT of integration by parts.
Edit 3: You'll know you're at the end of integration by parts when you reach an relatively easily divisible polynomial fraction in the integral. If you need help going through the steps, let me know.
I agree you need integration by parts at some point, but I don't see how it's as complicated as you seem to think it is.
For the second problem, for one thing, once you expand the polynomial in partial fractions, you end up with the following 3 terms inside the integral:
one term that is trivial to integrate,
another that is possible to integrate with use of a table of trig derivatives,
and another term that requires one straightforward application of integration by parts.
As for the first problem, the only significant part of the problem is clever choice of u and dv (besides knowing the derivative of inverse sine). One application of integration by parts and a bit of algebra yields the solution immediately.
I have the answer for the first excercise:
Ok, change "sen" for "sin" (at begining and end, sorry for the math words in spanish)
I don't know if the answer is correct, I need practice. . . maybe it's too late for you
I couldn't finish the second one . . . I have to close a little business . . . see ya!
Ok, change "sen" for "sin" (at begining and end, sorry for the math words in spanish)
I don't know if the answer is correct, I need practice. . . maybe it's too late for you
I couldn't finish the second one . . . I have to close a little business . . . see ya!
funny how you mixed spanish and english there
BTW I SUCK at math...
_________________
One of God's own prototypes. Some kind of high powered mutant never even considered for mass production. Too weird to live, and too rare to die.
Eh, no prob. These aren't easy integrals by any stretch, though they do look more intimidating than they should. Plus you're right - it's so easy to get out of practice with such problems, since knowing how to solve integrals like these is 100% knowing what "trick" to use, and it is hard to keep such tricks memorized for a long time if you don't use them.
And, you never really get truly comfortable and fluent with these types of problems in the intro calculus courses. It wasn't until I took a lot of more advanced courses that used a lot of calculus that I really got fluent in solving integrals like these.
for one class, I had a teaching assistant from argentina who mixed spanish and english math. Didn't help that he wrote the letters weird also (some of the letters he wrote were sideways... )
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