Question

How would you go about evaluating the integral of (3x-1)/(4-x^2)^1/2 , let x = 2 sin theta ?

Answer 1

" x = 2 sin theta" is a little bit wtf. Are these supposed to be limits of integration? Where does theta come from? What are we integrating with respect to?

Answer 2

No limits of integration, I believe it's supposed to be a trig substitution problem but I'm having trouble with the mechanics of it.

Answer 3

If it clears anything up, the problem says let x = 2 sin theta

Answer 4

Okay I see it. Just a moment.

Answer 5

\[3(2sint) \div \sqrt{(4-(2sint)^2)}\]
\[6sint \div \sqrt{(4-4sint^2)}\]
\[6sint \div \sqrt{(4*(1-sint^2)}\]

Answer 6

\[6sint \div 2\sqrt{1-sint^2}\]
Now do a U substitution win u = sint^2

Answer 7

no, u = sint. Not sint^2

Answer 8

Let me know if that helps. Ask any questions you have.

Answer 9

http://www.wolframalpha.com/input/?i=%283x-1%29%2F%284-x%5E2%29%5E1%2F2+
there is a step by step here if you are still lost

Answer 10

I get to \[(6\sin \theta -1)\div(2\cos \theta)\] and tried to split the integral into \[3\sin \theta \div \cos \theta\] and \[-1 \div 2\cos \theta\] but I get stuck there. Am I on the right track?

Answer 11

Not sure how you get to 6sint - 1... Doesn't it just simplify to 6sint?

Answer 12

\[6sint \div 2 cost = 3 tant\]

Answer 13

3x -1 from the original integral with x = 2 sin theta = 6sin theta -1, no?

Answer 14

Ooh. Yes yes.

Answer 15

Yes. You're on the right track... 2sint/cost is easy to solve with a u substitution of u = cost.

Answer 16

3sint/cost which gets me to -ln cos t, but then I'm still left with 1/(2cos t), which I'm not sure how to antiderive

Answer 17

Okay haha this problem ends up being a fluttering nightmare the way it's been presented...

Answer 18

I can show you how the 3sint/cost part works out, but the 1/2cost is not pretty at all, and makes me wonder if we're interpreting this wrong somehow.

Answer 19

It should be relatively reasonable since it was given on an in class test, but I don't see where I could be going wrong so far.

Answer 20

Oh my. I figured it out.

Answer 21

Okay, let me point out that this integral is originally taken with respect to x. Since you've substituted in for x, you need to change the integral so that it is with respect to theta. This is how that is done:
\[x = 2sint \rightarrow dx/dt = 2cost \rightarrow dx = 2cost*dt\]

Answer 22

Everything else you did was correct though, up to the point:
\[\int\limits 6sint-1 \div 2cost dx\]

Answer 23

So now you just need to substitute dx = 2costdt to get:
\[\int\limits (6sint-1 \div 2cost) 2costdt\]

Answer 24

Which obviously simplifies to
\[\int\limits 6sint-1 dt\]

Answer 25

yodude?

Answer 26

That makes so much sense, can't believe I didn't think of it:P thanks man!

Answer 27

Haha yup. That sneaky change of variables.