Question

Integrate the function\[\int\limits_{}^{}\frac{ 1 }{ t ^{2}\sqrt{2-t ^{2}} }\]

Answer 1

Have you learned integration by parts yet?

Answer 2

yes

Answer 3

is that what i should use?

Answer 4

I think so, I haven't worked it out yet... but we can try that together if you like.
So in this example what would you choose for your "u" and which for your "dv"?

Answer 5

u would be \[\frac{ 1 }{ t ^{2} }\]
dv would be \[\frac{ 1 }{ \sqrt{2-t ^{2}} }\]

Answer 6

keep in mind that we need to be able to integrate our "dv".

Answer 7

so i suppose we want to switch those

Answer 8

\[\frac{ 1 }{ \sqrt{2-t ^{2}} } \] is not easy to integrate.

Answer 9

right

Answer 10

so lets switch them.
If dv=t^-2 then what is v?
If u=the rest what is du?

Answer 11

for the record, I got the answer using a trig substitution
it may be doable with integration by parts, too

Answer 12

trig subtstitution? please explain

Answer 13

have you done tirg subs yet?

Answer 14

trig*

Answer 15

i've done trigonometric integrals

Answer 16

oh wait yes i have

Answer 17

\[\int{dt\over t^2\sqrt{2-t^2}}=\frac1{\sqrt2}\int{dt\over t^2\sqrt{1-(\frac t{\sqrt2})^2}}\]now let\[\frac t{\sqrt2}=\sin\theta\implies dt=\sqrt2\cos\theta d\theta\]

Answer 18

The integration by parts becomes pretty messy stick with Turing.

Answer 19

it usually does when you have both variable in the denom...

Answer 20

do you have any idea what to do next @MATTW20 ?

Answer 21

how did you get to dt=sqrt2 cos theta

Answer 22

\[\frac t{\sqrt2}=\sin\theta\\t=\sqrt2\sin\theta\\dt=\sqrt2\cos\theta d\theta\]

Answer 23

i see. needed that

Answer 24

now you substitute

Answer 25

right, and don't forget to sub for t^2

Answer 26

what am i substituting in for there?

Answer 27

I gave you a formula for t in the above 3 lines, you should be able to get t^2 from that

Answer 28

oh i see

Answer 29

so 2sin^2 theta

Answer 30

oh for t^2, yes

Answer 31

so then\[\frac{ \sqrt{2}\cos \theta }{ 2\sin ^{2}\theta \sqrt{1-\sin^2\theta } }\]

Answer 32

yes, and don't forget we still have a factor of \[\frac1{\sqrt2}\]outside the integral, so that cancels with the one on top

Answer 33

ok

Answer 34

simplify as much as possible, what do you get?

Answer 35

1/2sin^2 theta?

Answer 36

right, integral of which is...?

Answer 37

might want to use one more trig identity before you integrate...

Answer 38

take out t^2 common out of root and substitute 1/t^2=c

Answer 39

-1/2tan theta

Answer 40

cot...

Answer 41

right

Answer 42

do you know how to get back to terms of t?

Answer 43

this is going to sound really bad but no

Answer 44

no, it's one of the more bizarre parts of trig subs if you ask me...
you have to find the triangle that corresponds to your initial substitution

Answer 45

so int this case our initial sub was\[\frac t{\sqrt2}=\sin\theta\]from which we can draw a triangle

Answer 46

oh i have that actually

Answer 47

right, so for you, u=t, and a=?

Answer 48

sqrt 2

Answer 49

yup, so cot(theta)=?

Answer 50

sqrt(2-t^2)/t^2

Answer 51

great, and so our final solution is...?

Answer 52

well what are we doing now?

Answer 53

putting back the -1/2 factor that we have out in front and putting a +C is all

Answer 54

oh yeah i forgot about that -1/2

Answer 55

alright cool

Answer 56

cool, hope you followed that well enough to tackle the rest on your own :)

Answer 57

that was actually a pretty complicated problem imo

Answer 58

trig subs seem really complex at first, but once you get used to them they are pretty formulaic

Answer 59

ummmmm what should i look for next time because i didn't think it was that obvious it was a trig substitution problem

Answer 60

when you have \[\frac1{\sqrt{a^2\pm x^2}}\] trig sub should come to mind

that said, there may have been a really smart sub we could have used to get the same answer in an easier way, but this is what came to me
by parts would not work though, and finding the right u-sub to make an integral like this easy is often as tricky as just going ahead with the trig sub, so it's a matter of what you are more comfortable with and stuff

Answer 61

ok thnx a lot

Answer 62

welcome!