Question

Change limits of integration for a trig substitution?

What needs to be done if the limits of integration are not in the domain of the arcsin function? I'm integrating (2-y^2)^3/2dy from -1 to 2. My trig sub is y=sqrt(2)*sin(theta), so to find theta I get sin^(-1)(y/sqrt(2)) is not in the reals when y=2.

How is this to be handled? Thank you very much for any help!!!

Answer 1

Hmm... maybe just use the substitution to find the indefinite integral, change back to the original function, and then enter the limits?

Answer 2

Did that but I end up with a theta in the answer, which can only be found using sin^(-1) so I can't evaluate it.

Answer 3

Do you have to use trig substitution? It wouldn't be hard to expand out the cubic.

Answer 4

For full credit, yes. Teacher is very good but a bit of a "stickler". If I just plug it into a formula and get anything at all wrong I get zero credit for the problem, and this problem is worth the most points on the entire quiz. (thanks for your help so far!)

Answer 5

Okay.. I'll try using the trig substitution and see where it goes/if it can be done without needing to enter the limits. If you do end up with a theta in the indefinite integral, then... well idk how you'd deal with that.

Answer 6

Right, that's exactly my problem, is not knowing how handle the limits of integration. I end up with \[\sin^{-1} \left( \frac{ 2 }{ \sqrt{2}} \right)\] which is not a real number.

Answer 7

ignoring the limits \[\Large \frac{ 1 }{ 2 } \int\limits (2-y^2)^3 dy \]
letting \[ \large y=\sqrt2 \sin \theta\]

Answer 8

Eventually I get to \[\large 4 \sqrt 2 \int\limits \cos^7 \theta d \theta\]
And it looks like it'll be solvable with a triangle.

Answer 9

Note that this trig substitution just seems to make this integral much more complex than just expanding out the original cubic :/

Answer 10

I'll finish it off later when I have time.