anonymous
  • anonymous
I need to understand this problem. Integrate 7xcos^4x^2 dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits_{}^{}7xcos ^{4}x ^{2} dx\]
anonymous
  • anonymous
Allright. First we need to take care of that x^2 inside the cosine.
anonymous
  • anonymous
Do you see how we would do that?

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anonymous
  • anonymous
By using a u sub?
anonymous
  • anonymous
Good job :) .
anonymous
  • anonymous
So u=x^2 and du=2x dx
anonymous
  • anonymous
or du/2 = x dx
anonymous
  • anonymous
Recall that 7 is a constant which can be pulled outside the intergal.
anonymous
  • anonymous
So what do you obtain?
zepdrix
  • 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
  • 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
  • anonymous
Yep that's what I am doing @zepdrix :P . Just walking him through :) .
anonymous
  • anonymous
her* :P
anonymous
  • anonymous
du/2 is really 1/2 * du . We can also pull out that 1/2 in front of the integral :) .
anonymous
  • anonymous
Sorry X) .
anonymous
  • anonymous
so we CAN use the cosine reduction formula?
anonymous
  • anonymous
Yep :) .
anonymous
  • anonymous
We obtain: |dw:1378616587529:dw|
anonymous
  • anonymous
|dw:1378616604878:dw|
anonymous
  • anonymous
Apply your half angle formula for cosine :) .
zepdrix
  • zepdrix
That's not the reduction formula .. +_+
anonymous
  • anonymous
I didn't use reduction :P .
anonymous
  • anonymous
I used the half angle formula.
anonymous
  • anonymous
|dw:1378616732667:dw|
zepdrix
  • zepdrix
Oh, she said "we CAN", i interpretted that as, "can we take that route". my mistake :3
anonymous
  • anonymous
I'm just more familiar with the reduction formula due to my problems earlier
anonymous
  • anonymous
I'm not >.< .
anonymous
  • anonymous
But do you notice that's the same thing as: |dw:1378616856756:dw|
anonymous
  • anonymous
|dw:1378616886323:dw|
anonymous
  • anonymous
I'm just more familiar with the reduction formula due to my problems earlier
anonymous
  • anonymous
@zepdrix : wanna take over then? I am not too familiar with the reduction formulas since I never used them.
anonymous
  • anonymous
I keep getting kicked off, sorry
anonymous
  • anonymous
Sorry @JPeg16 : I am used to the half angle formulas :/ .
zepdrix
  • zepdrix
Ya wio tried using using half angles to explain it earlier, i think we should try a different method :c
anonymous
  • anonymous
I'll go ahead and type out the solution
anonymous
  • anonymous
wio drove me crazy
zepdrix
  • zepdrix
Lol :)
anonymous
  • anonymous
|dw:1378617001415:dw|
zepdrix
  • zepdrix
Woops that should be cos(2u) at this point right? :o
anonymous
  • anonymous
No...
zepdrix
  • zepdrix
\[\Large \cos^2u \quad=\quad \frac{1}{2}(1+\cos2u)\]
anonymous
  • anonymous
Yeah. Whoops >.< .
anonymous
  • anonymous
that's what I thought
zepdrix
  • zepdrix
dem silly double angles D:>
anonymous
  • anonymous
lol
zepdrix
  • zepdrix
I didn't mean to hijack yer question dids +_+ I just remember how to get through this with redux :D
anonymous
  • anonymous
|dw:1378617197208:dw|
zepdrix
  • zepdrix
oh if you wanna leave the steps with double, that seems fine c:
anonymous
  • anonymous
I would like to see both ways
anonymous
  • anonymous
then use double angles again?
zepdrix
  • zepdrix
last term is cos^2(2u) right? XD just making sure hehe
anonymous
  • anonymous
Right >.> .
anonymous
  • anonymous
|dw:1378617286544:dw|
anonymous
  • anonymous
|dw:1378617300718:dw|
anonymous
  • anonymous
|dw:1378617339638:dw|
anonymous
  • anonymous
So the second cos is cos 4u?
anonymous
  • anonymous
|dw:1378617384855:dw|
anonymous
  • anonymous
Yeah. Basically we add a 1 along and multiply the inside angle by 2 more or less.
anonymous
  • anonymous
And divide the result by a 2.
anonymous
  • anonymous
oh, because that had the formula applied twice to it
anonymous
  • anonymous
|dw:1378617488758:dw|
anonymous
  • anonymous
Yeah :P .
anonymous
  • anonymous
|dw:1378617534114:dw|
anonymous
  • anonymous
And there we are :) .
zepdrix
  • zepdrix
Yayyy good times! Time for the Redux Show??? \:D/
anonymous
  • anonymous
And don't worry @zepdrix you didn't hijack the question :) .
anonymous
  • anonymous
Thank you @Dido525. I have less experience with that formula so it was very helpful :)
zepdrix
  • 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
  • anonymous
@wio : I didn't know he asked it before :P .
zepdrix
  • 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
  • zepdrix
This method look a little more familiar with Jennifer Peggy? :U
zepdrix
  • zepdrix
familiar maybe*
zepdrix
  • zepdrix
oh lol, when i posted the general formula, that should be (n-1)/n, not 3/4. silly me -_- but whatever
anonymous
  • anonymous
yep, I wasn't having luck with @wio teaching it though
anonymous
  • anonymous
You weren't doing any work though.
anonymous
  • 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
  • zepdrix
So I guess you wanna distribute the 7/2 from that step. :o
anonymous
  • anonymous
you weren't explaining anything. I wouldn't have asked the question if I knew what to do
zepdrix
  • zepdrix
oh snap +_+ we gotsa fight brewin
anonymous
  • anonymous
and my name is Jenny, not Jennifer :P
zepdrix
  • zepdrix
lies
zepdrix
  • zepdrix
dat peggles -_-
anonymous
  • anonymous
@zepdrix How did you get to 99? I'm still stuck at 87.
anonymous
  • anonymous
I have been stuck at 67 for almost a year and a half now.
zepdrix
  • zepdrix
Hmm, my `questions answered` is a bit larger than yours. I guess I just spend too much time on here lol.
anonymous
  • anonymous
You have a 50% medal and I have 25%
anonymous
  • anonymous
I'm getting robbed.
zepdrix
  • zepdrix
50%? whu? :u

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