Please, please help with substitution rule for indefinite integrals:

- anonymous

Please, please help with substitution rule for indefinite integrals:

- jamiebookeater

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- anonymous

\[\int\limits_{}^{} (4z + \sin(3z))e ^{6z ^{2}-\cos(3z)} dz\]

- anonymous

whew, that's a doozy.

- anonymous

So what do you think you want to use for substitution?

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## More answers

- anonymous

\[\int\limits_{?}^{?} \sin(x) d x = -\cos(x) + constant\]

- anonymous

Not sure...possibly something to do with 3z? I have no idea to be honest

- anonymous

I would let u = that yucky exponent on the e

- anonymous

Ignore the two ?'s

- anonymous

Okay but how does that help?

- anonymous

Well.
\[u = 6x^2 - cos(3z)\]
\[\implies du = 12z +3sin(3z)dz = 3(4z + sin(3z))dz\]

- anonymous

Oh so then \[du = 2z ^{3}-\sin(3z)dx\]

- anonymous

Close, but you integrated. You want to take the derivative

- anonymous

oh

- anonymous

Okay. Where do I plug u into?

- anonymous

Well first rewrite dz in terms of du using that last equation I had.

- anonymous

?

- anonymous

\(du = 3(4z + sin(3z))dz \implies dz =\ ?\)

- anonymous

I don't understand - what am I solving for?

- anonymous

Well for this part you're solving for dz.

- anonymous

Err...I divide 3(4z + sin(3z)) on both sides?

- anonymous

Yes

- anonymous

So \[dz = \frac{1}{3(4z+sin(3z))} du\]

- 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

Wait a minute...Why do I replace dz?

- 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

We are replacing our nasty expression with z for a nice expression with u.

- anonymous

\[(4z + \sin(3z)e ^{u}\]du

- anonymous

But remember that dz does not equal du.

- anonymous

ButI thought we replaced dz

- anonymous

\[dz = \frac{1}{3(4z+sin(3z))} du\]

- anonymous

So you have to put that whole thing in for dz

- anonymous

I'm so lost

- anonymous

Lets try an easy one to start with.

- anonymous

\[\int x*e^{x^2}dx\]

- anonymous

We let \(u = x^2\)

- anonymous

So \[du = (2x)dx \implies dx = \frac{1}{2x}du\]

- anonymous

With me so far?

- anonymous

sure

- anonymous

If not, please ask

- anonymous

yes

- anonymous

Ok. So then
\[\int x*e^{x^2}dx = \int x*e^u(\frac{1}{2x} du)\]

- anonymous

And we can cancel the x in front with the x in the denominator from the du.

- anonymous

to get
\[\frac{1}{2}\int e^udu\]

- anonymous

Which is nice and easy.

- anonymous

\[\frac{1}{2}\int e^udu = \frac{1}{2}e^u + c = \frac{1}{2}e^{x^2} + C\]

- anonymous

So I've got\[1/3 e ^{u}\] for my problem
when do I integrate?

- anonymous

Now.

- anonymous

Before or after I plug in u

- anonymous

Don't switch back your u until after you integrate.

- anonymous

How do I integrate that?

- anonymous

\[\int \frac{1}{3}e^u du = \frac{1}{3}\int e^udu\]

- anonymous

What is the integral of \(e^x\)?

- anonymous

e^x

- anonymous

Right.

- anonymous

\[1/3 *e ^{6z ^{2}-\cos(3z)} \]

- anonymous

+C

- anonymous

oh right

- anonymous

Nice job!

- anonymous

Oh my goodness. Would you please walk me through one more?

- anonymous

Sure, but you'll have to do more of it this time ;)

- anonymous

How's it start?

- anonymous

Okay I'll try:

- anonymous

\[\int\limits_{}^{}(4z ^{3} + 6\csc ^{2}(6z)(z ^{4}-\cot(6z))^{4} dz\]

- anonymous

Ugh

- anonymous

Do I need to FOIL first?

- anonymous

No.

- anonymous

These are all ugly u substitution problems. Foiling will just make you cry.

- anonymous

You want to find a nice thing to pick for a u substitution.

- anonymous

Oh okay

- anonymous

Do you notice anything interesting about these two things?

- anonymous

Not really sure what to look for, actually.

- 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

Err expression, not equation

- anonymous

Oh! I guess the 4z^3 and z^4

- anonymous

And what about the \(6csc^2(6z)\) and the -cot(6z) ?

- anonymous

AND the csc^2

- anonymous

right.

- anonymous

So what you want to do is pick the one that is not the derivative, and let that be u.

- anonymous

So let\[u = z ^{4} - \cot(6z)\]

- anonymous

Yep, then solve for dz in terms of z and du.

- anonymous

So \[\int\limits_{}^{}(4z ^{3}+6\csc ^{2}(6z))u ^{4}du\]

- anonymous

Nope. but close.

- anonymous

\[u = z^4 - cot(6z)\]
So
\[du =\ ?\]

- anonymous

4z^3 + 6csc^2(6z)

- anonymous

close.

- anonymous

\[du = (4z^3 + 6csc^2(6z))dz\]

- anonymous

So then what does dz = ?

- anonymous

1 over all that mess

- anonymous

yep.

- 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

\[(4z ^{3} + 6\csc ^{2}(6z))u ^{4} * 1/4z ^{3} + 6\csc ^{2}(6z)\]

- anonymous

simplify!

- anonymous

u^4?

- anonymous

yes

- anonymous

\[(z ^{4} - \cot(6z))^{4}\]

- anonymous

No

- 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

? I don't know

- anonymous

I can't pull anything out in front of the integral sign

- anonymous

\[\int x^2dx = \frac{x^3}{3} +C\]
Right?

- anonymous

Oh! Integrate!

- anonymous

Yes. Once you have something nice, you just integrate.

- anonymous

Then after you integrate you plug back in what u was.

- anonymous

\[(z ^{4} - \cot(6z)^{5}/ 5 +c\]

- anonymous

Yep.

- anonymous

WOW. Good grief. Thanks so much. I am going to use these as examples for the next ones. Thanks again

- anonymous

Very welcome =)

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