Question

How would you integrate (x^2)( sqrt of (1+(x^3)) )?

Answer 1

oh, i meant that x^2 is multiplied to the the second term.

Answer 2

um du= 3x^2

Answer 3

Oh not division? :O my bad

Answer 4

it's okay; i appreciate your help in typing out the steps

Answer 5

:)

Answer 6

\[\Large\bf\sf \int\limits \sqrt{\color{royalblue}{1+x^3}}\left(\color{orangered}{x^2\;dx}\right)\]

Answer 7

\[\Large\bf\sf \color{royalblue}{u=1+x^3}\]\[\Large\bf\sf du=3\color{orangered}{x^2\;dx}\]

Answer 8

Ok good, see how our du ALMOST matches up with the stuff in the brackets?

Answer 9

yes.

Answer 10

\[\large \int\limits_{}^{} x^2 \sqrt{1+x^3}~dx\]
\[\large \int\limits_{}^{} \sqrt{1+x^3}~x^2dx\]
\[u = 1 + x^3\]
\[du = 3x^2 dx\]
\[\frac{ 1 }{ 3 } du = x^2 dx\]
Set it up with u-sub now.

Answer 11

...

Answer 12

is it \[\sqrt{1+x ^{3}} (1/3)du\]

Answer 13

Before integrating?

Answer 14

You replaced the differential dx correctly.
We need to replace the x also though,\[\Large\bf\sf \int\limits\limits \sqrt{\color{royalblue}{1+x^3}}\left(\color{orangered}{x^2\;dx}\right)\quad =\quad \int\limits\limits \sqrt{\color{royalblue}{u}}\left(\color{orangered}{\frac{1}{3}\;du}\right)\]

Answer 15

you can factor out the constant (1/3)

Answer 16

\[\Large\bf\sf \frac{1}{3}\int\limits u^{1/2}\;du\]Just apply the normal Power Rule from there! :)

Answer 17

zepdrix, where did you get the blue sqrt of u from? sorry if it's a dumb question

Answer 18

That was our initial thing that we chose to be u,\[\Large\bf\sf \color{royalblue}{u=1+x^3}\]The substitution we chose^

Answer 19

ohh

Answer 20

okay i will apply the upper limit of 2 and lower limit of 0 now

Answer 21

after i integrate.

Answer 22

thank you

Answer 23

np.

Keep in mind that those limits of integration apply to x! Not u.
After integrating, you will need to `undo` your substitution before you plug those values in.

Answer 24

oh how do i undo it?

Answer 25

do i replace the u with 1+x^3?

Answer 26

Yes, after you integrate, before plugging in your 0 and 2.
Put 1+x^3 back in place of your u.

Answer 27

thx

Answer 28

um how would i use u substitution for How do you integrate [(cosx)^4] (sinx)?

Answer 29

i tried u=cosx but got the wrong answer in the end

Answer 30

nvm i got it