Question

Help substituting with indefinite integrals? http://puu.sh/8chRE.jpg

Answer 1

Did you have any initial thoughts on what you might try to substitute? We are typically looking for a substitution whose derivative is attached to the dx.

Answer 2

* Oh, I should mention that the link is actually an indefinite integral, you might have meant to post a different one?

Answer 3

I misstyped it. and we would sub (16-x^3) with u correct?

Answer 4

which would make it int
x^2/u^2dx

Answer 5

Well, almost. When you do a u-substitution, you want to substitute du as well as u. The integration is done with respect to the variable in d(), so if you have both variables you can't really figure out which one to work with.
To find du, we take its derivative: \(\dfrac{d}{dx} u = \dfrac{d}{dx} \left(16 - x^3 \right) \)
And then multiply the dx to both sides: \( du = \dfrac{d}{dx} \left(16 - x^3 \right) \ dx \)

Answer 6

\( \displaystyle \int \frac{x^2}{(16 - x^3)^2} \ dx \)
dx acts a lot like multiplication, we can put it up on top of the numerator:
\( \displaystyle \int \frac{\color{blue}{x^2 \ dx}}{(\color{green}{16 - x^3})^2} \)
\( u = \color{green}{16 - x^3} \)
\( du = \dfrac{d}{dx} \left(16 - x^3 \right) \ dx \qquad du= -3 \color{blue}{x^2 dx} \)
Can you see the idea? We want to replace the blue part with something in du as we replace the green part with u.

Answer 7

will it be -1/3du/u^2?

Answer 8

Yes, there you go. :)
Now the integral is done with respect to u. It may help to rewrite the u into the numerator by using negative exponents. :)

Answer 9

\( \displaystyle \int \frac{ -\dfrac{1}{3} \ du}{u^2} \) You can pull constant multiples out of integrals. Then the u^2 in the denominator can come up because 1/u^2 = u^(-2). That let's us use the power rule for integrals.

Answer 10

so -1/3 du / (u^3/3)

Answer 11

Not quite. I mean to say:
\( \displaystyle -\frac{1}{3} \int \frac{du}{u^2} = - \frac{1}{3} \int u^{-2} \ du \)

The power rule on integrals: \( \displaystyle \int u^n \ du = \frac{u^{n+1}}{n+1} +C\)
Note that n= -2 in our case.

Answer 12

\( \displaystyle \color{gray}{- \frac{1}{3}} \int u^{\color{blue}{-2}} \ du = \color{gray}{-\frac{1}{3}} \dfrac{u^{\color{blue}{-2}+1}}{\color{blue}{-2}+1} \color{gray}{+C} \)

Just trying to highlight the important parts there. If you understand that part, we just have to substitute back in our \( u = 16 - x^3 \) directly.

Answer 13

Or if any part is unclear, feel free to ask! I can try my best to explain, or you can tag someone else as well to get another perspective on explaining it. :)