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
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
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
Ok so you can write the integral as this:
$\displaystyle I(x)=\frac{x^3}{x^2+16}=\frac{x^3+16x-16x}{x^2+16}=\frac{x(x^2+16)}{x^2+16}-\frac{16x}{x^2+16}= x-\frac{16x}{x^2+16}$
$\displaystyle
\int{I(x)} = \int {xdx}-\int {\frac{16x}{x^2+16}}$
Put $\displaystyle t = x^2+16$
So,
$\displaystyle dt = 2xdx$ and $\displaystyle 8dt = 16xdx$
Substitute all this,
$\displaystyle I=\frac{x^2}{2}-\int{\frac{8dt}{t}}$
which gives
$\displaystyle I=\frac{x^2}{2}-8ln|t|+c $
back substitute t to get
$\displaystyle I=\frac{x^2}{2}-8ln|x^2+16|+c$
which is your answer, no trigonometry required!
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?
$\displaystyle I = \int {\frac{x^3}{\sqrt{x^2+16}}}dx$
Plug in $\displaystyle t^2 = x^2+16$ to get
$\displaystyle 2tdt=2xdx$ or
$\displaystyle tdt=xdx$
This changes your integrand to
$\displaystyle I=\int {\frac{(t^2-16)tdt}{t}}$
which reduces, quite simply, to
$\displaystyle I=\int {(t^2-16)dt}$
$\displaystyle I=t^3/3-16t+c$
back substitute t^2
$\displaystyle Answer=(\sqrt{(x^2+16)})^3-16\sqrt{x^2+16}+c$
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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Don't integrate - balloontegrate!
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