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How would I begin to solve this integrand ? by completing the square ?
I try completing the square when I factor out a - 4 outside the square root but it does not work out quite right.
This one is just a tad confusing to me when trying to complete the square, which I would assume would be the best way to do this.
Note thatQuote:
Originally Posted by The Power
How would I begin to solve this integrand ? by completing the square ?
I try completing the square when I factor out a - 4 outside the square root but it does not work out quite right.
This one is just a tad confusing to me when trying to complete the square, which I would assume would be the best way to do this.
.
So, I was going in the general direction correctly by using completing the square method ?
Hi,
I dont really know where fantastic is going with his tip but the problem looks like an arcsin/arccos to me,
the definition is:
derivative of arcsin(x) is:
http://upload.wikimedia.org/math/1/3...97221020b7.png
derivative of arccos(x) is:
http://upload.wikimedia.org/math/b/7...6f6cbbc8b4.png
Make it look like 1 + something^2 in the square root and the solution should not be far away.
I think you meant "...1 - something^2...", and this is what I believe fantastic was heading to: he already got that the expression within the square root is (sqrt(5))^2 - (2x - 3)^2 = 5 {1 -[(2x-3)/Sqrt(5)]^2}, and if we get this into the square root in the denominator this is exactly an Arcsin integral.Quote:
Originally Posted by knivmannen
Tonio
Where that last bit comes from ? I see where 5 comes from, but notQuote:
Originally Posted by mr fantastic
(2x - 3)^2
Quote:
Originally Posted by The Power
First, check that 5 - (2x - 3)^2 really equals what it must; second, in fantastic's post, just before he obtained this expression, he took out (1/4) from the parentheses but here he got;
[5/4 - (x - 3/2)^2] = [5/4 - ((2x - 3)/2)^2] = [5/4 - (1/4)(2x-3)^2] and now 1/4 goes out and etc.
Tonio
Still dont see where 1/4 and 2x magically comes from
Ok, this already is high school algebra, so "get your hands dirty" and begin working on it (at the very least grab a pencil and a sheet of paper and write down both fantastic's message and mine to see clearly what's written there...and then begin doing maths!)Quote:
Originally Posted by The Power
Tonio
Ah, I see, he expanded out (x - 3/2) ^2 the multiplied, through -4 so techincally his simplifaction isn't needed just more easier.
Quote:
Originally Posted by The Power
I don't know what he did, but no need at all to expand anything:
(x - 3/2)^2 = ([2x-3]/2)^2 = (1/4)*(2x-3)^2
Tonio
Do I use the trig. substitution of arcsin x ?
Quote:
Originally Posted by The Power
I supsoe you can try substitution, but in fact:
(1/5)[1/Sqrt(1 - [(2x-3)/Sqrt(5)]^2] =
(1/2Sqrt(5)))[(2/Sqrt(5))/Sqrt(1 - [(2x-3)/Sqrt(5)]^2] ==>
the integral of the above equals (1/2Sqrt(5))*Arcsin[(2x-3)/sqrt(5)]
Tonio
Mind throwing some math tags around those ?
Couldn't I use 2x - 3 = sin theta and solve for x, I tried that but its seems to complicate things
tonio has graciously walked you through basic algebra that a student attempting questions like the one you posted should have had no trouble with. So if you want math tags, I suggest you quote what tonio wrote and then insert the tags yourself into the quote. And the least you could do is to actually acknowledge the help you got by saying thankyou (or clicking on the Thanks button).Quote:
Originally Posted by The Power
If you follow what's been posted in this thread then it should be clear that you have a standard form. I had hoped the basic substitution u = 2x - 3 would have made that obvious. I suggest you go back and review this material.