1. ## Trig Substitution

The problem is the integral of (x^3)/(((x^2)+16)^.5) dx x=4tan(theta)
I've worked it down to 4^2 times the integral of (tan^3(theta))/(sec(theta)) dtheta, but don't know what to do with it. Any help is appreciated.

Thanks

2. $\frac{x^3}{x^2 + 16} = \frac{x(x^2 + 16) - 16x}{x^2 + 16} =$ (split the numerator)

(The give-away is the higher power top than bottom - or does it have to be trig?)

3. My apologies, I have been up all night and am feeling stupid. it's actually the square root of x^2 + 16 on bottom which is what requires the trig.

4. Still the x^3 on top means you don't need trig... use - Wolfram|Alpha and click on show steps, but I'll post a picture, just in case it helps

5. Originally Posted by Hellacious D
The problem is the integral of (x^3)/((x^2)+16) dx x=4tan(theta)
I've worked it down to 4^2 times the integral of (tan^3(theta))/(sec(theta)) dtheta, but don't know what to do with it. Any help is appreciated.

Thanks
Ok so you can write the integral as this:

$I(x)=\frac{x^3}{x^2+16}=\frac{x^3+16x-16x}{x^2+16}=\frac{x(x^2+16)}{x^2+16}-\frac{16x}{x^2+16}= x-\frac{16x}{x^2+16}$
$
\int{I(x)} = \int {xdx}-\int {\frac{16x}{x^2+16}}$

Put $t = x^2+16$
So,

$dt = 2xdx$ and $8dt = 16xdx$

Substitute all this,

$I=\frac{x^2}{2}-\int{\frac{8dt}{t}}$
which gives
$I=\frac{x^2}{2}-8ln|t|+c$

back substitute t to get

$I=\frac{x^2}{2}-8ln|x^2+16|+c$

6. Tom, I appreciate your pointing out of that tool for checking answers, but the way in which it gets answers is nothing like how I'm being taught and doesn't appear to be something I can replicate. tanujkush, sorry for wasting your time. I realized I had put x^2 + 16 and meant to put ((x^2)+16)^.5 on bottom. It has been fixed now.

7. Originally Posted by Hellacious D
I've worked it down to 4^2 times the integral of (tan^3(theta))/(sec(theta)) dtheta
Whoops... multiply this by the derivative of 4 tan theta, though. Pic coming

8. Originally Posted by Hellacious D
Tom, I appreciate your pointing out of that tool for checking answers, but the way in which it gets answers is nothing like how I'm being taught and doesn't appear to be something I can replicate. tanujkush, sorry for wasting your time. I realized I had put x^2 + 16 and meant to put ((x^2)+16)^.5 on bottom. It has been fixed now.

Oh then its even more simple,

so your integrand is this i hope?

$I = \int {\frac{x^3}{\sqrt{x^2+16}}}dx$

Plug in $t^2 = x^2+16$ to get
$2tdt=2xdx$ or
$tdt=xdx$

$I=\int {\frac{(t^2-16)tdt}{t}}$

which reduces, quite simply, to

$I=\int {(t^2-16)dt}$

$I=t^3/3-16t+c$

back substitute t^2

$Answer=(\sqrt{(x^2+16)})^3-16\sqrt{x^2+16}+c$

9. Should also work with trig, but since you're putting another function inside x you need to multiply what you had by the derivative of the inner function, according to the chain rule, which I like to depict so...

Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite and hence subject to the chain rule).

So, the problem so far...

The chain rule also the key to the new integrand ...

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