Question

I have a test tmrw and I have NO CLUE how to solve these integrals! example:

integrate dx/x(sqrt(x^2+16))

and

integrate x^3(sqrt(9-x^2))dx

can someone give me some help?

Answer 1

Oooo trig stuff :O fun!

Answer 2

\[\int \frac{dx}{x(\sqrt{x^2+16})}\] I believe this follows one of the basic integral forms of \(\sqrt{a^2+u^2}\)\[\sf \int \frac{du}{u\sqrt{u^2+a^2}} = -\frac{1}{a}\ln \left|\frac{\sqrt{a^2+u^2}}{u}\right|+C\]

Answer 3

Am I supposed to be seeing an equation? sorry.

on the second one I got

Answer 4

integral u(sqrt(9-u))du where u=x^2 and 1/2du=xdx

Answer 5

Hmm I don't think that u-sub is going to be very effective.
If you insist on doing a u-sub, you'll want to do \(\large\rm u=9-x^2\)
But trig sub would be a lot better :O

Answer 6

hmm.. thats part of it. \[\int x^3\sqrt{9-x^2}dx \qquad \implies \qquad \int\color{blue}{ x}\cdot \color{red}{x^2}\sqrt{9-\color{red}{x^2}}\color{blue}{dx}\]\[\color{red}{u=x^2}~,~ \color{blue}{du=2xdx \implies xdx =\dfrac{1}{2}du} \]

Answer 7

zepdrix (sorry Jhannybean) can you give me a push as to what a should be for the second problem?

Answer 8

Bahhh I take back what I said :)
The u-sub I mentioned should work out really nicely!
\(\large\rm u=9-x^2\)

Answer 9

Ah ha... I just read up on the trig subs. \[\sqrt{9-x^2} \qquad \implies\qquad \sqrt{a^2-x^2} ~,~ x=a\sin(\theta)\]This!!! Now I recall.

Answer 10

You do that because you really would like to get rid of the subtraction under the root.
\(\large\rm du=-2x~dx\qquad\to\qquad -\frac{1}{2}du=x~dx\)
But also:\[\large\rm u=9-x^2\qquad\to\qquad x^2=u+9\]
And then sub all your pieces in! :)\[\large\rm \int\limits x^2\sqrt{9-x^2}~(x~dx)\]

Answer 11

Oh boy... Can anybody teach me how to set up the "triangle" to know to use cos, sin or tan?

Answer 12

Therefore \[\int x^3\sqrt{9-x^2}dx \qquad \implies \qquad \int(3\sin(\theta))^3\sqrt{9-(3\sin(\theta))^2 }\cdot 3\cos(\theta)d\theta\] Something like that?...

Answer 13

Where \(x=3\sin(\theta) ~,\qquad ~ dx = 3\cos(\theta)d\theta\)

Answer 14

ok... I only know how to do a trig subs with fractions where dx is on top...

Answer 15

Wow,yeah I agree @zepdrix , this is going to be really long x_x haha crap.

Answer 16

I'll group the steps together that I posted earlier,
maybe it will help :o

Answer 17

\[\large\rm \int\limits\limits x^3\sqrt{9-x^2}~dx=\int\limits \color{#F35633}{x^2}\sqrt{\color{#CC0033}{9-x^2}}~(\color{green}{x~dx})\]Making the substitution:\[\large\rm \color{#CC0033 }{u=9-x^2}\]Adding x^2 to each side, subtracting u from each side gives us,\[\large\rm \color{#F35633}{x^2=9-u}\]Differentiating the red one gives us,\[\large\rm du=-2x~dx\qquad\to\qquad \color{green}{-\frac{1}{2}du=dx}\]\[\large\rm \int\limits\limits \color{#F35633}{x^2}\sqrt{\color{#CC0033}{9-x^2}}~(\color{green}{x~dx})=\int\limits\limits \color{#F35633}{(9-u)}\sqrt{\color{#CC0033}{u}}~\left(\color{green}{-\frac{1}{2}du}\right)\]

Answer 18

From there it's not too bad, ya? :)
Rewrite your square root as a 1/2 power,\[\large\rm =-\frac{1}{2}(9-u)u^{1/2}~du\]Distribute the u^(1/2),
and then just apply your power rule to each term.

Answer 19

Doh :d typo on the green step, my bad\[\large\rm du=-2x~dx\qquad\to\qquad \color{green}{-\frac{1}{2}du=x~dx}\]

Answer 20

HEY! What if I set u^2=9-x^2? because then the sqrt is resolved to just u, and I'm left with

integrate: (9-u^2)(-u^2)du?

Answer 21

That's a really strange way to approach it :D
But I think it works!
Let's see...\[\large\rm u^2=9-x^2,\qquad\quad x^2=u^2+9,\qquad\quad u=\sqrt{9-x^2}\]Differentiating the first part,\[\large\rm 2u~du=-2x~dx\qquad\to\qquad -u~du=x~dx\]Subbing everything in gives you,\[\large\rm \int\limits x^2\sqrt{9-x^2}~(x~dx)=\int\limits (u^2+9)\cdot u\cdot (-u~du)\]I think maybe your sign is off in the middle, looks good besides that though.

Answer 22

Naw your sign is good :) mine was bad lol

Answer 23

I worked it out both waym and they both check with the answer in the book! Any help on the first one? I know it has to do with x=sina or something?

Answer 24

\[\large\rm \int\limits \frac{dx}{x \sqrt{x^2+16}}\]You would really like to get \(\large\rm (stuff)^2+1\) under the root, then you could make the substitution:\[\large\rm stuff=\tan\theta\]And we would get,\[\large\rm (\tan \theta)^2+1=\sec^2\theta\]Trig sub is a clever way to get rid of the addition/subtraction between terms!
It allows us to easily take this out of the root.

Answer 25

How do we get +1 though?
Hmm we have a +16...

Well, if we can make a 16 show up in our "stuff",
then we can factor a 16 out of each term, leaving us with 16(stuff^2+1) under the root.

Answer 26

Hopefully that makes some sense :p
I know I'm throwing the word stuff around a lot...

Answer 27

So it seems that sine isn't the sub we're looking for.
Since we have `addition`, we're going to head over to tangent,
since the `square identity` involving tangent has addition in it.

Answer 28

Looking around, I found this helpful powerpiont. It gives a good start to anyone looking at this thread when its archived:
https://drive.google.com/file/d/0B0lXunoGraQaQ2tfZ0NCbFpUeDQ/edit

Answer 29

Speaking of, we need an a(tan)theta = x for this one.

Answer 30

lol

Answer 31

give me a second to work this out a bit...

Answer 32

Did you work it out? I like this question so I'll show a solution you can then compare your answer to..
\[\int\limits \frac{ dx }{ x \sqrt{x^2+16} }\] \[x= \tan \theta \implies dx = \sec^2\theta d \theta\] that will be our substitution, so our new integral will look like \[\int\limits \frac{ 4\sec^2\theta d \theta }{ \tan \theta \sqrt{(4\tan \theta)^2+16} }\] now notice in the denominator we can factor out the 16, \[\int\limits \frac{ 4 \sec^2 \theta d \theta }{ 4\tan \theta \sqrt{16 \tan^2 \theta+16} } \implies \int\limits \frac{ \sec^2 \theta d \theta }{ 4 \tan \theta \sqrt{16(\tan^2 \theta+1)} }\] now we can use the following identity \[\tan^2 \theta +1 = \sec^2 \theta\] so we get \[\int\limits \frac{ \sec^2 \theta d \theta }{ \tan \theta \sqrt{16 \sec^2 \theta} } \implies \frac{ 1 }{ 4 }\int\limits \frac{ \sec^2 \theta }{ \tan \theta \sec \theta }d \theta\]\[\frac{ 1 }{ 4 } \int\limits \frac{ \sec \theta d \theta }{ \tan \theta } d \theta \implies \frac{ 1 }{ 4 } \int\limits \csc \theta d \theta\] and remember the integral of csc theta is \[\frac{ 1 }{ 4 } \int\limits \csc \theta d \theta = \frac{ 1 }{ 4 } \ln|\csc \theta-\cot \theta|+C\] but we're not exactly done yet as we made a substitution earlier. looking at the trig ratios we finally get \[\frac{ 1 }{ 4 }(\ln|x|-\ln|\sqrt{x^2+16}+4|)+C\]