Question

Which trigonometric substitution would you use in evaluating the following integral?
Integral (x^2 − 49)^(−3/2) dx

choices : http://i.imgur.com/8cDNL.png

Can someone go through how to approach this?
I missed the class today so I'm sure what to do.
I thought it will be 7 tan u but it was wrong

Answer 1

*not sure

Answer 2

Hey!
So we need to remember a couple of trig identities
\[\cos^2(\theta)=1-\sin^2(\theta)=1-[\sin(\theta)]^2=1-x^2 \text{ if I let } \sin(\theta)=x\]
\[\sec^2(\theta)=1+tan^2(\theta)=1+[tan(\theta)]^2=1+x^2 \text{ I if I let } tan(\theta)=x\]
\[tan^2(\theta)=\sec^2(\theta)-1 =[\sec(\theta)]^2-1=x^2-1 \text{ if I let } \sec(\theta)=x\]

Answer 3

So you see what substitution you will need based on what I have here?

Answer 4

Like we can kinda play with these a little

Answer 5

Or we could play with your problem...
\[\int\limits_{}^{}(x^2-49)^\frac{3}{2} dx\]
\[(49)^\frac{3}{2} \int\limits_{}^{}(\frac{x^2}{49}-1)^\frac{3}{2} dx\]
\[(49)^\frac{3}{2} \int\limits_{}^{}(\frac{x}{7})^2-1)^\frac{3}{2} dx\]
So try letting \[\text{ using } \sec(\theta)=f(x) \text{ based on the thingys I wrote \above }\]

Answer 6

And I'm missing an open paranthesis but I think you see that I was....

Answer 7

oh so x = 7 sec (theta)
Thanks for the explanation :)

Answer 8

Yep yep! :)
You are awesome! :)