Question

Use a change of variables to evaluate the following definite integral.

Answer 1

So I've rearranged things a tiny bit.
\[\Large \int\limits_{x=0}^3\frac{1}{(\color{orangered}{4x^2+36})^2}\left(\color{royalblue}{x\;dx}\right)\]
We want to make the substitution,\[\Large \color{orangered}{u=4x^2+36}\]What would we get for our \(\Large du\) ?

Answer 2

do we differentiate it? if so du= 8x dx?

Answer 3

Yes very good.\[\Large du=8\color{royalblue}{x\;dx}\]See how we have an x dx in our problem?
It'd be nice if we could replace it with something involving du.
But we have this 8 in the way, what can we do? Hmm

Answer 4

ah! (1/8)du = xdx, then subs to the (xdx) in the expression?

Answer 5

yes good :)

Answer 6

\[\Large \color{royalblue}{\frac{1}{8}du=x\;dx}\]

Answer 7

\[\Large \int\limits\limits_{x=0}^3\frac{1}{(\color{orangered}{4x^2+36})^2}\left(\color{royalblue}{x\;dx}\right)\qquad\to\qquad \int\limits\limits_{x=0}^3\frac{1}{(\color{orangered}{u})^2}\left(\color{royalblue}{\frac{1}{8}du}\right)\]There is one thing to be careful about though.
Our limits of integration are still `in terms of x`.

We have a couple of options, depending on which you're more comfortable with:
~We can now come up with new bounds for u based on our substitution,
~Or we can integrate, and then undo the substitution and plug the x bounds in.

Answer 8

i think im more comfortable with integrating first then undo-ing, so the integral would be -1/u, then just undo right then sub in the limits, i think i got it

Answer 9

k cool :)

Answer 10

thanks! (: