Question

quick trig substitution question, what trig identity do I use to substitute in for sqrt(x^2+a^2)? I know sqrt(a^2+x^2)=atan(theta)

Answer 1

for example

Answer 2

You know that? Good.
So it likes like our a value is 6

Answer 3

so just 6tan(theta)

Answer 4

Yes, very good.
That takes care of the denominator nicely.
\[\Large\rm x=6\tan \theta\]\[\Large\rm x^2=36\tan^2\theta\]
\[\Large\rm \int\limits \frac{x^3}{\sqrt{36+x^2}}dx=\int \frac{x^3}{\sqrt{36+36tan^2\theta}}dx\]\[\Large\rm \int\limits \frac{x^3}{6\sqrt{1+\tan^2\theta}}dx=\frac{1}{6}\int\limits \frac{x^3}{\sqrt{\sec^2\theta}}dx\]

Answer 5

ok so does the sec^2(theta) just turn into sec(theta)

Answer 6

or a better question how do i get rid of this square root

Answer 7

Good good good.\[\Large\rm\frac{1}{6}\int\limits\limits \frac{\color{orangered}{x^3}}{\sec\theta}\color{royalblue}{dx}\]Then just a few other pieces that you have to apply your substitution to.

Answer 8

That was the whole point of making a trig sub, our identity GOT RID OF the addition between the terms.
Made it a whole lot easier to just take the root.

Answer 9

1/6 int 216tan^3(theta)/sec(theta) -----> 36 int tan^3(theta)/sec(theta)

Answer 10

oh and i need to add the dx.

Answer 11

1/6 int 216tan^3(theta)*6sec^2(x)/sec(theta)

Answer 12

\[\Large\rm\frac{216}{6}\int\limits\limits\limits \frac{\color{orangered}{\tan^3\theta}}{\sec\theta}\color{royalblue}{6\sec^2\theta}\]Mmm ok looks good so far.

Answer 13

216 int tan^3(theta) sec(theta)

Answer 14

Cool.
From there, if you can't seem to figure it out right away,
a good thing to try is converting everything to sines and cosines.
Sometimes you get things canceling out nicely.
I dunno if we need to do that here though, one sec >.< Thinking..

Answer 15

You could do this I guess.\[\Large\rm \tan^3\theta \sec \theta=\color{#CC0033}{\tan^2\theta}\tan \theta \sec \theta=\color{#CC0033}{(\sec^2\theta-1)}\tan \theta \sec \theta\]And then \(\Large\rm u=\sec\theta\)

Answer 16

If that little trick is difficult to see though,
then maybe sines and cosines is the way to go.

Answer 17

yeah it makes sense with the trying to get the exponent even. i still need to memorize the identities like tan^2(theta)--> sec^2(theta-1)

not quite sure what i do with the u though..

Answer 18

The derivative of sec(theta)=sec(theta)tan(theta).
That one popped into my head.

Answer 19

\[\Large\rm 216\int\limits (\color{#F35633}{\sec}^2\color{#F35633}{\theta}-1)\color{#3399AA}{(\sec \theta \tan \theta d \theta)}\]So we've got,\[\Large\rm \color{#F35633}{u=\sec \theta}\]And taking the derivative gives us,\[\Large\rm \color{#3399AA}{du=\sec \theta \tan \theta d \theta}\]

Answer 20

ok i think i got it from here. thanks zep

the color coding was really helpful

Answer 21

The ending can be a little tricky sometimes.
You remember how to `undo` your substitution using triangle math?

Answer 22

i wouldn't mind you walking me through it (:

Answer 23

\[\Large\rm x=6\tan \theta\qquad\to\qquad \tan \theta=\frac{x}{6}=\frac{opposite}{adjacent}\]

Answer 24

It looks like after you integrate, you're going to end up with a bunch of secants.
So you'll need to know the hypotenuse length.

Answer 25

We just use the Pythagorean Theorem for that, yes?