anonymous
  • anonymous
Please, please help with substitution rule for indefinite integrals:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{}^{} (4z + \sin(3z))e ^{6z ^{2}-\cos(3z)} dz\]
anonymous
  • anonymous
whew, that's a doozy.
anonymous
  • anonymous
So what do you think you want to use for substitution?

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anonymous
  • anonymous
\[\int\limits_{?}^{?} \sin(x) d x = -\cos(x) + constant\]
anonymous
  • anonymous
Not sure...possibly something to do with 3z? I have no idea to be honest
anonymous
  • anonymous
I would let u = that yucky exponent on the e
anonymous
  • anonymous
Ignore the two ?'s
anonymous
  • anonymous
Okay but how does that help?
anonymous
  • anonymous
Well. \[u = 6x^2 - cos(3z)\] \[\implies du = 12z +3sin(3z)dz = 3(4z + sin(3z))dz\]
anonymous
  • anonymous
Oh so then \[du = 2z ^{3}-\sin(3z)dx\]
anonymous
  • anonymous
Close, but you integrated. You want to take the derivative
anonymous
  • anonymous
oh
anonymous
  • anonymous
Okay. Where do I plug u into?
anonymous
  • anonymous
Well first rewrite dz in terms of du using that last equation I had.
anonymous
  • anonymous
?
anonymous
  • anonymous
\(du = 3(4z + sin(3z))dz \implies dz =\ ?\)
anonymous
  • anonymous
I don't understand - what am I solving for?
anonymous
  • anonymous
Well for this part you're solving for dz.
anonymous
  • anonymous
Err...I divide 3(4z + sin(3z)) on both sides?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
So \[dz = \frac{1}{3(4z+sin(3z))} du\]
anonymous
  • anonymous
So now. Go back to your integral, and replace the exponent of e with u, and the dz with your expression for du, and you'll be pleasantly surprised.
anonymous
  • anonymous
Wait a minute...Why do I replace dz?
anonymous
  • anonymous
Because we are no longer integrating with respect to z, we are integrating with respect to u now. Because it turns out u is nicer.
anonymous
  • anonymous
We are replacing our nasty expression with z for a nice expression with u.
anonymous
  • anonymous
\[(4z + \sin(3z)e ^{u}\]du
anonymous
  • anonymous
But remember that dz does not equal du.
anonymous
  • anonymous
ButI thought we replaced dz
anonymous
  • anonymous
\[dz = \frac{1}{3(4z+sin(3z))} du\]
anonymous
  • anonymous
So you have to put that whole thing in for dz
anonymous
  • anonymous
I'm so lost
anonymous
  • anonymous
Lets try an easy one to start with.
anonymous
  • anonymous
\[\int x*e^{x^2}dx\]
anonymous
  • anonymous
We let \(u = x^2\)
anonymous
  • anonymous
So \[du = (2x)dx \implies dx = \frac{1}{2x}du\]
anonymous
  • anonymous
With me so far?
anonymous
  • anonymous
sure
anonymous
  • anonymous
If not, please ask
anonymous
  • anonymous
yes
anonymous
  • anonymous
Ok. So then \[\int x*e^{x^2}dx = \int x*e^u(\frac{1}{2x} du)\]
anonymous
  • anonymous
And we can cancel the x in front with the x in the denominator from the du.
anonymous
  • anonymous
to get \[\frac{1}{2}\int e^udu\]
anonymous
  • anonymous
Which is nice and easy.
anonymous
  • anonymous
\[\frac{1}{2}\int e^udu = \frac{1}{2}e^u + c = \frac{1}{2}e^{x^2} + C\]
anonymous
  • anonymous
So I've got\[1/3 e ^{u}\] for my problem when do I integrate?
anonymous
  • anonymous
Now.
anonymous
  • anonymous
Before or after I plug in u
anonymous
  • anonymous
Don't switch back your u until after you integrate.
anonymous
  • anonymous
How do I integrate that?
anonymous
  • anonymous
\[\int \frac{1}{3}e^u du = \frac{1}{3}\int e^udu\]
anonymous
  • anonymous
What is the integral of \(e^x\)?
anonymous
  • anonymous
e^x
anonymous
  • anonymous
Right.
anonymous
  • anonymous
\[1/3 *e ^{6z ^{2}-\cos(3z)} \]
anonymous
  • anonymous
+C
anonymous
  • anonymous
oh right
anonymous
  • anonymous
Nice job!
anonymous
  • anonymous
Oh my goodness. Would you please walk me through one more?
anonymous
  • anonymous
Sure, but you'll have to do more of it this time ;)
anonymous
  • anonymous
How's it start?
anonymous
  • anonymous
Okay I'll try:
anonymous
  • anonymous
\[\int\limits_{}^{}(4z ^{3} + 6\csc ^{2}(6z)(z ^{4}-\cot(6z))^{4} dz\]
anonymous
  • anonymous
Ugh
anonymous
  • anonymous
Do I need to FOIL first?
anonymous
  • anonymous
No.
anonymous
  • anonymous
These are all ugly u substitution problems. Foiling will just make you cry.
anonymous
  • anonymous
You want to find a nice thing to pick for a u substitution.
anonymous
  • anonymous
Oh okay
anonymous
  • anonymous
Do you notice anything interesting about these two things?
anonymous
  • anonymous
Not really sure what to look for, actually.
anonymous
  • anonymous
\((4z^3 + 6csc^2(6z)) * (z^4 - cot(6z))^4\) You are looking for something in the equation that looks like the derivative of another part of the equation.
anonymous
  • anonymous
Err expression, not equation
anonymous
  • anonymous
Oh! I guess the 4z^3 and z^4
anonymous
  • anonymous
And what about the \(6csc^2(6z)\) and the -cot(6z) ?
anonymous
  • anonymous
AND the csc^2
anonymous
  • anonymous
right.
anonymous
  • anonymous
So what you want to do is pick the one that is not the derivative, and let that be u.
anonymous
  • anonymous
So let\[u = z ^{4} - \cot(6z)\]
anonymous
  • anonymous
Yep, then solve for dz in terms of z and du.
anonymous
  • anonymous
So \[\int\limits_{}^{}(4z ^{3}+6\csc ^{2}(6z))u ^{4}du\]
anonymous
  • anonymous
Nope. but close.
anonymous
  • anonymous
\[u = z^4 - cot(6z)\] So \[du =\ ?\]
anonymous
  • anonymous
4z^3 + 6csc^2(6z)
anonymous
  • anonymous
close.
anonymous
  • anonymous
\[du = (4z^3 + 6csc^2(6z))dz\]
anonymous
  • anonymous
So then what does dz = ?
anonymous
  • anonymous
1 over all that mess
anonymous
  • anonymous
yep.
anonymous
  • anonymous
So plug in 1 over all that mess times du for dz in the original equation. and plug in u for what we said it equaled.
anonymous
  • anonymous
\[(4z ^{3} + 6\csc ^{2}(6z))u ^{4} * 1/4z ^{3} + 6\csc ^{2}(6z)\]
anonymous
  • anonymous
simplify!
anonymous
  • anonymous
u^4?
anonymous
  • anonymous
yes
anonymous
  • anonymous
\[(z ^{4} - \cot(6z))^{4}\]
anonymous
  • anonymous
No
anonymous
  • anonymous
\[\int(4z^3 + 6csc^2(6z)) * (z^4 - cot(6z))^4dz\] \[ = \int (4z^3 + 6csc^2(6z)) u^4(\frac{1}{4z^3 + 6csc^2(6z)})du \] \[= \int u^4du =\ ?\]
anonymous
  • anonymous
? I don't know
anonymous
  • anonymous
I can't pull anything out in front of the integral sign
anonymous
  • anonymous
\[\int x^2dx = \frac{x^3}{3} +C\] Right?
anonymous
  • anonymous
Oh! Integrate!
anonymous
  • anonymous
Yes. Once you have something nice, you just integrate.
anonymous
  • anonymous
Then after you integrate you plug back in what u was.
anonymous
  • anonymous
\[(z ^{4} - \cot(6z)^{5}/ 5 +c\]
anonymous
  • anonymous
Yep.
anonymous
  • anonymous
WOW. Good grief. Thanks so much. I am going to use these as examples for the next ones. Thanks again
anonymous
  • anonymous
Very welcome =)

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