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.
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:
derivative of arccos(x) is:
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.
Tonio
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
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).
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.