Question

Anyone good with integrals?

Answer 1

integral of (16-x^2)^-2

Answer 2

Have you learned about Trigonometric Substitutions?
That seems like the most straight-forward approach here.

Answer 3

yes I believe it is asin(theta)?

Answer 4

\[\Large\rm \int\limits\limits \frac{1}{(16-x^2)^2}~dx\]

Mmm yes good. Understand why the 4 is here?\[\Large\rm x=4 \sin \theta\]

Answer 5

yes, I don't know what to do after I sub in 4sin(theta) for the x

Answer 6

\[\Large\rm \int\limits\frac{1}{(16-16\sin^2\theta)^2}~dx\]Factor out a 16 from each term,\[\Large\rm =\int\limits\limits\frac{1}{16^2(1-\sin^2\theta)^2}~dx\]Is that step ok?

Answer 7

yes you do, and so I have 1/[256cos^4(theta)] right now and cant go further

Answer 8

Mmm ok looks good so far.
We need to replace our dx with something involving d(theta) before we can integrate.

Answer 9

\[\Large\rm x=4 \sin \theta\]\[\Large\rm dx=?\]

Answer 10

4cos(theta)

Answer 11

\[\Large\rm dx=4\cos \theta ~d \theta\]Ok good, plugging it in gives us...

Answer 12

\[\Large\rm =\int\limits\limits\limits\frac{1}{4^4\cos^4\theta}~dx=\int\limits\limits\limits\frac{1}{4^4\cos^4\theta}~(4\cos \theta~d \theta)\]

Answer 13

I rewrote 16^2 as 4^4, I hope that wasn't confusing.
It will make things easier to cancel out.

Answer 14

yes I am following so far

Answer 15

Umm so we can cancel some stuff out right?\[\Large\rm \frac{1}{4^3}\int\limits \frac{1}{\cos^3\theta}~d \theta=\frac{1}{4^3}\int\limits \sec^3\theta~ d \theta\]

Answer 16

do you apply a reduction formula next?

Answer 17

Mmmm yah good call.
We have an `odd power` of secant.
So I don't think we have a nice clean substitution.

Answer 18

ok so I have \[\sec \theta \tan \theta \over 2\] + (1/2)ln|sectheta+tantheta| +C

Answer 19

Here is our reduction formula:
\[\Large\rm \mathcal I_n=\frac{1}{n-1}\sec^{n-2}\theta \tan \theta+\frac{n-2}{n-1}\mathcal I_{n-2}\]Where,\[\Large\rm I_n=\int\limits \sec^n \theta ~d \theta\]

Answer 20

Oh you figured it out already hah XD nice.

Answer 21

Mmmm ok great, I think that looks correct so far!

Answer 22

ok good, now we have to change the thetas to x's?

Answer 23

\[\Large\rm \frac{1}{4^3}\left[\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\ln\left|\sec \theta+ \tan \theta\right|\right]+c\]Yes we need this back in terms of x.

Answer 24

To do so, we need to go back to triangle trig.

Answer 25

\[\Large\rm x=4\sin \theta \qquad\to\qquad \sin \theta=\frac{opposite}{hypotenuse}=\frac{x}{4}\]

Answer 26

Understand how to change back to x? :O

Answer 27

yes I do I got \[2x \over 16-x^2 \] \[+ 1/2\ln|(4+x)/\sqrt{16-x^2}| \]
is this correct?

Answer 28

So you canceled a 2 in the numerator with the 1/2 in front of the first term?
Umm looks good.

But don't forgot about your 1/4^3 that we were carrying around for a long time.

Answer 29

so all of this is multiplied my 1/64? also if I want this integral from 0 to 5 what would that turn into...im getting a sqrt(-9) on the bottom and I don't think that is correct

Answer 30

and yes I did cancel a 2 and the 1/2

Answer 31

The 5 is causing a problem? Hmmmm...

Answer 32

unless im doing something wrong?

Answer 33

Was that part of the problem?
Or did you just pull the 0 and 5 out of thin air? XD

Answer 34

no lol it is part of the problem

Answer 35

actually, sorry it supposed to be from 0 to 2. my bad

Answer 36

Oh lol, ok that would make more sense :)

Answer 37

ok so final answer is 1/3 + 1/2ln(6/sqrt(12)) all times 1/64?

Answer 38

Mmmm I simplified a little differently.. lemme see if mine matches yours heh.

Answer 39

Mmm ok yes very good!

Answer 40

Here is way you can check your work if it will help.
https://www.wolframalpha.com/input/?i=integral+from+x%3D0+to+x%3D2+of+1%2F%2816-x%5E2%29%5E2

If you plug your result into your calculator, you should get the same decimal value that Wolfram is producing.

Answer 41

ok multiplying through I get 1/192 + 1/128ln|6/(2sqrt(3))|

Answer 42

Good.
Might be a nice idea to `rationalize` the inside of the log.

Answer 43

how do you do that?

Answer 44

\[\Large\rm \frac{6}{2\sqrt3}=\frac{3}{\sqrt3}\]Multiply top and bottom by sqrt3,\[\Large\rm \frac{3\sqrt3}{\sqrt3 \sqrt3}=\frac{3\sqrt3}{3}\]Which simplifies a little further yes?

Answer 45

By rationalize what I mean is, get the irrational value out of the denominator.
We don't like to leave ugly square roots in the bottom like that.

Answer 46

oh ok got it. also, are the absolute bars necessary or should I change it to parens?

Answer 47

Yah dropping them seems fine.

Answer 48

ok thank you soooo much!!

Answer 49

np \c:/ yay team!