Understanding Trigonometric Substitution for Evaluating Integrals

[kuahji]

Evaluate

\int1/sqrt(4x^2-49) 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

\int8dx/(4x^2+1)^2 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.

 

[rock.freak667]

If you take 2x=7sec\theta then:

2 dx=7 sec\theta tan\theta \theta

\int \frac{1}{\sqrt{4x^2-49}} dx

\equiv \int \frac{\frac{7}{2}sec \theta tan \theta}{\sqrt{49sec^2 \theta-49}

then use sec^2\theta-1=tan^2 \theta

for the 2nd integral. they use tan because when you substitute x=1/2 tanu then the denomination becomes

(4(1/2 tanu)^2 +1)^2 = (tan^2u+1)^2 and well tan^u+1=sec^2 u
so it becomes sec^4u.

But basically what you want to do is make a substitution where the denominator (after substitution) can be used as another trig identity. e.g sin^2u+cos^2=1 etc.

(EDIT: Not sure if my LaTex is showing up correctly[My browser is showing LaTex from questions I typed out many days ago!] so I don't know if you will understand what I wanted to say)

 

[kuahji]

Yes, it makes sense now. Its usually the little stuff that gets me all confused ^_^. If you take 2x=7sec is what I wasn't getting.