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#1 2012-03-05 19:49:50
- Kryptonis
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Integration with Inverse Trig function
I'm having a bit of a time with this. I have tried integration by parts and just can't seem to get it right....
∫(3-3x)/(sqrt(64-9x^2)) dx
#2 2012-03-05 19:58:43
- anonimnystefy
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Re: Integration with Inverse Trig function
Hi
Use the substitution u=8*sin(x)/3
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#3 2012-03-07 13:46:24
- Kryptonis
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Re: Integration with Inverse Trig function
Thanx a bunch, was totally going in the wrong direction on this
#4 2012-03-07 20:02:35
- anonimnystefy
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Re: Integration with Inverse Trig function
If there's anything else you don't understand,ask freely over here.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#5 2012-03-31 18:11:54
- bob bundy
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Re: Integration with Inverse Trig function
hi brabenderjack
Welcome to the forum!
Speaking personally, I don't think there is a guaranteed way of doing integration. With differentiation, once you can do the basic functions (powers, trig, exp, log) and you can do product, quotient and chain rule, you can differentiate anything made up from these.
But there's no such approach for integration.
Sometimes it is 'obvious' that it is a certain function just by looking and remembering the rules for differentiation.
Sometimes it 'looks' like integration by parts may work.
If I see the presence of a function and its derivative then I know substitution will help.
Some other problems also 'yield' to substitution.
Beyond that, it's just trial and improvement for me.
Usually, for exams, the examiners only set ones that can be done by one of the above methods.
There are some fairly simple looking functions that won't give a 'nice' answer at all!
If I got one I could not do, I'd post it here! 
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#6 2012-03-31 21:47:05
- anonimnystefy
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Re: Integration with Inverse Trig function
Hi bob
here's one for you:
Last edited by anonimnystefy (2012-03-31 21:53:03)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#7 2012-04-01 00:28:52
- bob bundy
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Re: Integration with Inverse Trig function
hi Stefy
just a bit of 'light relief' before I go back to your sequence.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#8 2012-04-01 00:30:54
- anonimnystefy
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Re: Integration with Inverse Trig function
As i gathered from your LaTeX code that looks right,but there is a simpler form.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#9 2012-04-01 00:35:46
- bob bundy
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Re: Integration with Inverse Trig function
Latex corrected.
looking for a simplification.
B
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#10 2012-04-01 00:39:31
- bob bundy
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Re: Integration with Inverse Trig function
How about
?
B
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#11 2012-04-29 11:41:11
- anonimnystefy
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Re: Integration with Inverse Trig function
Where has e**x vanished?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón