anonymous
  • anonymous
Evaluate the intergal
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{}^{}3\sin^2(t)\cos^4(t)dt\]
anonymous
  • anonymous
My work: \[3\int\limits_{}^{}\sin^2(x)\cos^4(t)dt\] \[3\int\limits_{}^{}\sin^2(t)\cos^2(t)\cos^2(t)dt\] \[3\int\limits_{}^{}\frac{ 1 }{ 2 }(1-\cos(2t))(\frac{ 1 }{ 2 }(1+\cos(2t)))^2dt\] \[\frac{ 3 }{ 8 }\int\limits_{}^{}(1-\cos(2t))((1+\cos(2t)))^2dx\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}((\cos^3(2t)+\cos^2(2t)+\cos(2t)+1) dt\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}((cos^2(2t))(cos(2t))+\cos^2(2t)+\cos(2t)+1))dt\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}(\cos^3(2t)+\cos^2(2t)+\cos(2t)+1)dt\]
anonymous
  • anonymous
I don't know what to do after :( .

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anonymous
  • anonymous
The 4th step should be a dt.
anonymous
  • anonymous
look in the back of your book at the "reduction" formula i think you can make a u sub earlier on though. maybe i am wrong
anonymous
  • anonymous
Is there no other way?
anonymous
  • anonymous
I mean I did get a little further.
anonymous
  • anonymous
\[\frac{ -3 }{ 8 }\int\limits\limits_{}^{}(\cos^2(2t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]
anonymous
  • anonymous
i would start with \[\int\cos^4(x)dx-\int\cos^6(x)dx\] and then look in the back of the book
anonymous
  • anonymous
\[\frac{ -3 }{ 8 }\int\limits\limits\limits_{}^{}((1-\sin^2(t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]
anonymous
  • anonymous
Where did that come from?
anonymous
  • anonymous
there are fomulas that i cannot remember for integrating \(\sin^n(x)\) and \(\cos^n(x)\) and i garantee you they are on the back cover of your text
anonymous
  • anonymous
Yeah, I know. I turn them into the half angles, which I did.
anonymous
  • anonymous
That stupid cos^2(2t) is getting in the way.
anonymous
  • anonymous
\(\sin^2(x)=1-\cos^2(x)\) then multiply out it is easier than reinventing the wheel you are trying to derive the formula, (which is admirable) but it is easier to look it up
anonymous
  • anonymous
Using another half angle won't help because they are multiplied together which just makes it squared afain,
anonymous
  • anonymous
again*
anonymous
  • anonymous
why i find this topic rather dull. almost everything you need is printed on the back cover of your text
anonymous
  • anonymous
I know. I have to do it though.
anonymous
  • anonymous
then i really recommend looking up the formula for \(\int\cos^n(x)dx\)
anonymous
  • anonymous
So I get:\[\frac{ -3 }{ 8 }\int\limits_{}^{}(1-\sin^2(2t))\cos(2t)+(1-\sin^2(2t)+\cos(2t)+1)dt\]
anonymous
  • anonymous
Hmm good point.
anonymous
  • anonymous
Sec.
anonymous
  • anonymous
Can I ask a dumb question?
anonymous
  • anonymous
We have that constant out in front. If we break up the intergal into seperate sums do I distribute that constant to all the intergals?
anonymous
  • anonymous
sure so i can seem dumber if i cannot answer it
anonymous
  • anonymous
yes (whew)
anonymous
  • anonymous
ignore the annoying 3, put it in at the end
anonymous
  • anonymous
Hehe. Thanks. This should be easy now.
anonymous
  • anonymous
here it is \[\int\cos^n(x)dx = \frac{1}{n}\cos(x)\sin(x)+\frac{n-1}{n}\int \cos^{n-2}(x)dx\] use with \(n=4\) and again with \(n=6\)
anonymous
  • anonymous
Ohh that's MASSIVELY convenient.
anonymous
  • anonymous
it is going to be a pain, but much less of a pain than what you were doing
anonymous
  • anonymous
Why 4 and 6?
anonymous
  • anonymous
I have 3 and 2.
anonymous
  • anonymous
\[\sin^2(x)\cos^4(x)=(1-\cos^2(x))\cos^4(x)=\cos^4(x)-\cos^6(x)\]
anonymous
  • anonymous
so 4 and 6
anonymous
  • anonymous
Right.
anonymous
  • anonymous
Thanks!
anonymous
  • anonymous
have fun!
anonymous
  • anonymous
yw

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