I need to understand this problem. Integrate 7xcos^4x^2 dx

- anonymous

I need to understand this problem. Integrate 7xcos^4x^2 dx

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- anonymous

\[\int\limits_{}^{}7xcos ^{4}x ^{2} dx\]

- anonymous

Allright. First we need to take care of that x^2 inside the cosine.

- anonymous

Do you see how we would do that?

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## More answers

- anonymous

By using a u sub?

- anonymous

Good job :) .

- anonymous

So u=x^2 and du=2x dx

- anonymous

or du/2 = x dx

- anonymous

Recall that 7 is a constant which can be pulled outside the intergal.

- anonymous

So what do you obtain?

- zepdrix

After you get it into the `u form` like you had in the last post,
I would finish it up by using the `Cosine Reduction Formula` like we did earlier. :)
If applying double angle over and over works better for you though, you can do that.

- anonymous

okay, awesome. The 7 is part of what was throwing me off. So now we have \[7\int\limits_{}^{} \cos ^{4}(u) \frac{ du }{ 2 }\]

- anonymous

Yep that's what I am doing @zepdrix :P . Just walking him through :) .

- anonymous

her* :P

- anonymous

du/2 is really 1/2 * du . We can also pull out that 1/2 in front of the integral :) .

- anonymous

Sorry X) .

- anonymous

so we CAN use the cosine reduction formula?

- anonymous

Yep :) .

- anonymous

We obtain:
|dw:1378616587529:dw|

- anonymous

|dw:1378616604878:dw|

- anonymous

Apply your half angle formula for cosine :) .

- zepdrix

That's not the reduction formula .. +_+

- anonymous

I didn't use reduction :P .

- anonymous

I used the half angle formula.

- anonymous

|dw:1378616732667:dw|

- zepdrix

Oh, she said "we CAN", i interpretted that as, "can we take that route". my mistake :3

- anonymous

I'm just more familiar with the reduction formula due to my problems earlier

- anonymous

I'm not >.< .

- anonymous

But do you notice that's the same thing as:
|dw:1378616856756:dw|

- anonymous

|dw:1378616886323:dw|

- anonymous

I'm just more familiar with the reduction formula due to my problems earlier

- anonymous

@zepdrix : wanna take over then? I am not too familiar with the reduction formulas since I never used them.

- anonymous

I keep getting kicked off, sorry

- anonymous

Sorry @JPeg16 : I am used to the half angle formulas :/ .

- zepdrix

Ya wio tried using using half angles to explain it earlier, i think we should try a different method :c

- anonymous

I'll go ahead and type out the solution

- anonymous

wio drove me crazy

- zepdrix

Lol :)

- anonymous

|dw:1378617001415:dw|

- zepdrix

Woops that should be cos(2u) at this point right? :o

- anonymous

No...

- zepdrix

\[\Large \cos^2u \quad=\quad \frac{1}{2}(1+\cos2u)\]

- anonymous

Yeah. Whoops >.< .

- anonymous

that's what I thought

- zepdrix

dem silly double angles D:>

- anonymous

lol

- zepdrix

I didn't mean to hijack yer question dids +_+
I just remember how to get through this with redux :D

- anonymous

|dw:1378617197208:dw|

- zepdrix

oh if you wanna leave the steps with double, that seems fine c:

- anonymous

I would like to see both ways

- anonymous

then use double angles again?

- zepdrix

last term is cos^2(2u) right? XD
just making sure hehe

- anonymous

Right >.> .

- anonymous

|dw:1378617286544:dw|

- anonymous

|dw:1378617300718:dw|

- anonymous

|dw:1378617339638:dw|

- anonymous

So the second cos is cos 4u?

- anonymous

|dw:1378617384855:dw|

- anonymous

Yeah. Basically we add a 1 along and multiply the inside angle by 2 more or less.

- anonymous

And divide the result by a 2.

- anonymous

oh, because that had the formula applied twice to it

- anonymous

|dw:1378617488758:dw|

- anonymous

Yeah :P .

- anonymous

|dw:1378617534114:dw|

- anonymous

And there we are :) .

- zepdrix

Yayyy good times!
Time for the Redux Show??? \:D/

- anonymous

And don't worry @zepdrix you didn't hijack the question :) .

- anonymous

Thank you @Dido525. I have less experience with that formula so it was very helpful :)

- zepdrix

Recall:\[\Large I_n \quad=\quad \frac{1}{n}\sin x \cos^{n-1}x+\frac{3}{4}I_{n-2}\]
So we haveeeeee:\[\Large \frac{7}{2}\int\limits \cos^4u\;du \quad=\quad \frac{7}{2}I_4\]
\[\Large I_4 \quad=\quad \frac{1}{4}\sin x \cos^3x+\frac{3}{4}I_2\]

- anonymous

@wio : I didn't know he asked it before :P .

- zepdrix

\[\Large \frac{7}{2}I_4 \quad=\quad \frac{7}{2}\left(\frac{1}{4}\sin x \cos^3x+\frac{3}{4}I_2\right)\]

- zepdrix

This method look a little more familiar with Jennifer Peggy? :U

- zepdrix

familiar maybe*

- zepdrix

oh lol, when i posted the general formula, that should be (n-1)/n, not 3/4. silly me -_-
but whatever

- anonymous

yep, I wasn't having luck with @wio teaching it though

- anonymous

You weren't doing any work though.

- anonymous

yes, I love that formula now. It's more work in the short run but straight forward and less work in the long run

- zepdrix

So I guess you wanna distribute the 7/2 from that step. :o

- anonymous

you weren't explaining anything. I wouldn't have asked the question if I knew what to do

- zepdrix

oh snap +_+ we gotsa fight brewin

- anonymous

and my name is Jenny, not Jennifer :P

- zepdrix

lies

- zepdrix

dat peggles -_-

- anonymous

@zepdrix How did you get to 99? I'm still stuck at 87.

- anonymous

I have been stuck at 67 for almost a year and a half now.

- zepdrix

Hmm, my `questions answered` is a bit larger than yours.
I guess I just spend too much time on here lol.

- anonymous

You have a 50% medal and I have 25%

- anonymous

I'm getting robbed.

- zepdrix

50%? whu? :u

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