Question

∫(x^3)/(sqrt((x^2)+4)

Answer 1

I know you have to use trig substitution

Answer 2

\[\Large\bf\sf \frac{1}{2}\int\limits \frac{2x}{\sqrt{x^2+4}}\left(x^2\;dx\right)\]I would rewrite the expression like this and then do Integration by Parts.

Oh trig sub? Is that what you're learning about right now?
We can do that method instead.

Answer 3

yeah we are learning trig sub

Answer 4

Factor a 4 out of each term under the root,\[\Large\bf\sf \int\limits\limits \frac{x^3}{\sqrt{4\left[\left(x/2\right)^2+1\right]}}dx\]Then pull the the 4 out of the root,\[\Large\bf\sf \frac{1}{2}\int\limits\limits\limits \frac{x^3}{\sqrt{\left(\frac{x}{2}\right)^2+1}}dx\]This puts it into an easier form for us to set up a trig sub.

Answer 5

So trig sub will allow us to get rid of the addition under the root, making it easier to simplify.


We have (something)^2 + 1.
Which trig relationship should we use?
Do you remember which trig identity involves (something)^2 + 1 = (something else)^2

Answer 6

1+tan^2 =sec^2

Answer 7

Ok good good.
We'll want to use tangent.
That will allow us to simplify it down to secant^2 under the root.

So we're making the substitution:\[\Large\bf\sf \frac{x}{2}=\tan \theta\]

Answer 8

We'll need to replace the x in the numerator, and also the differential dx as well.

Answer 9

\[\Large\bf\sf \frac{1}{2}\int\limits\limits\limits\limits \frac{x^3}{\sqrt{\left(\color{orangered}{\frac{x}{2}}\right)^2+1}}dx\quad=\quad \frac{1}{2}\int\limits\limits\limits\limits \frac{x^3}{\sqrt{\left(\color{orangered}{\tan \theta}\right)^2+1}}dx\]

\[\Large\bf\sf =\quad \frac{1}{2}\int\limits\limits\limits\limits \frac{x^3}{\color{orangered}{\sec \theta}}dx\]Ok so we've dealt with the denominator so far.
That part make sense? :o

Answer 10

yep so far so good

Answer 11

So if this is our substitution, we can solve for x pretty easily:\[\Large\bf\sf \frac{x}{2}=\tan \theta\qquad\to\qquad x=2 \tan \theta\]What do you get for your differential?\[\Large\bf\sf dx=?\]

Answer 12

2(secx)^2

Answer 13

\[\Large\bf\sf dx=2\sec^2\theta\;d \theta\]Ok good.
So now we just plug all the pieces in and simplify.

Answer 14

\[\large\bf\sf =\quad \frac{1}{2}\int\limits\frac{(2\tan \theta)^3}{\sec \theta}\left(2\sec^2\theta\;d \theta\right)\quad = \quad 8\int\limits \frac{\tan^3 \theta}{\sec \theta}\left(\sec^2\theta\;d \theta\right)\]

Answer 15

Hopefully I plugged those in correctly :o

Answer 16

a u-sub will do nicely

Answer 17

so then it becomes 8∫(tanx)^3 * secx

Answer 18

so 8∫(tanx)^2(tanx)(secx)....then I got stuck

Answer 19

Umm ya that one is a little tricky.
I think one approach is to remember that: \(\Large\bf\sf (\sec x)'\quad=\quad \sec x\tan x\)
We might be able to use this. Hmm

Answer 20

Rewrite two of the tangent powers in terms of secant:\[\Large\bf\sf =8\int\limits \tan \theta (\sec^2\theta - 1)\sec \theta\;d \theta\]\[\Large\bf\sf =8\int\limits \sec^2\theta(\sec \theta \tan \theta)\;d \theta- 8\int\limits \sec \theta \tan \theta \;d \theta\]

Answer 21

how did you go from the first part to the second?

Answer 22

\[\Large\bf\sf =8\int\limits\limits \tan \theta \sec \theta(\sec^2\theta - 1)\;d \theta\]I distributed the tangent and secant to each term in the brackets.
Then wrote it as two separate integrals.

Answer 23

oh ok i see

Answer 24

\[\Large\bf\sf =8\int\limits\limits \sec^2\theta\color{orangered}{(\sec \theta \tan \theta\;d \theta)}- 8\int\limits\limits \color{orangered}{(\sec \theta \tan \theta\;d \theta)}\]Understand how to solve those integrals?
In each case, the orange part is the derivative of secant.

So in the first one, you can do a u-sub to make it a bit clearer.

Answer 25

so is is (8/3)(secx)^3 - 8secx

Answer 26

Ya looks good.
Then we need to get it back in terms of x, which can be a little tricky.
We need to go back to our good ole triangle math.