Question

find the integral of 1/x^2(sqrt(4-x^2) using trig substitution

Answer 1

I did try I ended up with a really complicated integral since I also substituted the 4sin^2theta for the x

Answer 2

2cos tdt=dx

Answer 3

4sin^2t=x^2

Answer 4

Yes

Answer 5

Wait did you ask me to try to do this function? @OOOPS ???

Answer 6

Yes

Answer 7

x = 2sec(t)
dx = 2sec(t)tan(t)
Integral ( [1/sqrt(4sec^2(t) - 4)] [2sec(t)tan(t)] )

Answer 8

Wait am I doing it right?

Answer 9

Integral ( [2sec(t)tan(t)] / [2tan(t)] )dt
Integral ( sec(t) ) dt

Answer 10

= ln |sec(t) + tan(t)| + C

Answer 11

That is all that I know how to do.

Answer 12

I think you should be using:\[x=2 \sin \theta\]

Answer 13

Umm I'm getting confused by all the replies a bit

Answer 14

Sorry.

Answer 15

Yes I made x that however x^2 confuses me since I end up with 1/4sin^2theta after the cancellations and I cannot past that

Answer 16

*get past I mean

Answer 17

The idea behind trig substitution is that you can turn something like:

\[\sqrt{1-\cos^2 \theta}\]
into
\[\sqrt{\sin^2 \theta}=\sin \theta\]
because

\[\sin^2 \theta + \cos^2 \theta = 1\]

and for your problem you have to change it by multiplying the whole thing by 4.
\[4\sin^2 \theta + 4 \cos^2 \theta = 4\]\[4-4\sin^2 \theta = 4 \cos^2 \theta\]

Answer 18

yes I get that under the square root and cancel it with the 2costheta dtheta generated in the numerator however the 4sin^2theta leftover for the x^2 in the denominator

Answer 19

Consider the following:
\[\frac{ 1 }{ \sin \theta }=\csc \theta\]

Answer 20

I don't understand how will this help me

Answer 21

Try plugging it in, maybe I'll remind you of one other thing... But not quite what you need... but almost...
\[\frac{ d }{ dx } \tan \theta = \sec^2 \theta\]

Answer 22

I guess I should have written: \[\frac{ d }{ d \theta}\] but you know.

Answer 23

I'm sorry I'm still a little confused

Answer 24

I would get integral csc^2theta which is cottheta this is still pretty complicated

Answer 25

cot tan theta

Answer 26

\[-\cot \theta = \int\limits_{}^{}\csc^2 \theta d \theta\]

That's pretty much all you need, not too complicated.

Answer 27

That isn't what the answer key says though it says -\[-\sqrt{4-x ^{2)}}/4x\]

Answer 28

You have to finish in terms of x of course.

Answer 29

I'll walk through the basic:
\[x=2\sin \theta\]
will give you:\[\int\limits_{}^{}\frac{ 2\cos \theta d \theta }{ 4\sin^2 \theta *2 \cos \theta }=\int\limits_{}^{}\csc^2\theta d \theta\]\[=-\frac{ 1 }{ 4 } \cot \theta\]

Now you need to plug back in for x!

Answer 30

Use the substitution you made.

Answer 31

So x/2 is sintheta and dx/2 is cos theta

Answer 32

and cottantheta is cos/sin

Answer 33

\[\frac{ -1 }{ 4 }\cot(\sin^{-1}(\frac{ x }{ 2 }))\]

That looks sloppy though, and besides it's simpler if we just think back to this:

\[\frac{ x }{ 2}=\sin \theta\] and now we remember SOH CAH TOA and draw this out:

Now when you do this::

\[\frac{ -1 }{4 }\cot(\theta)\]

just use that triangle, remember cot is just adjacent over opposite.

Answer 34

And oh, hey, it should match what your answer is. =)

Answer 35

Thank you that matches the answer. :)

Answer 36

Glad I could help, I love trig substitution along with pretty much all of calculus. =)