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[tex]\int1/sqrt(4x^2-49)[/tex] for x>7/2

Where I get lost really, is why do I set x = 7/2 sec u? The textbook just shows a generic formula where you always set x=a sec u. The only thing I could see is that anything less than 7/2 yiels a negative under the square root. But then again, this goes against the little formula which isn't really a problem, but take this other integral for example

[tex]\int8dx/(4x^2+1)^2[/tex] here it shows setting x=1/2 tan u. But here I'm not really understanding the reason why. I'm guessing its because I'm not really sure why I set x equal to say tanget, sine, or whatever else.

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# Homework Help: Trig Substituion Integral

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