Dido525
Evaluate the intergal
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Dido525
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\[\int\limits_{}^{}3\sin^2(t)\cos^4(t)dt\]
Dido525
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My work:
\[3\int\limits_{}^{}\sin^2(x)\cos^4(t)dt\]
\[3\int\limits_{}^{}\sin^2(t)\cos^2(t)\cos^2(t)dt\]
\[3\int\limits_{}^{}\frac{ 1 }{ 2 }(1-\cos(2t))(\frac{ 1 }{ 2 }(1+\cos(2t)))^2dt\]
\[\frac{ 3 }{ 8 }\int\limits_{}^{}(1-\cos(2t))((1+\cos(2t)))^2dx\]
\[\frac{ -3 }{ 8 }\int\limits_{}^{}((\cos^3(2t)+\cos^2(2t)+\cos(2t)+1) dt\]
\[\frac{ -3 }{ 8 }\int\limits_{}^{}((cos^2(2t))(cos(2t))+\cos^2(2t)+\cos(2t)+1))dt\]
\[\frac{ -3 }{ 8 }\int\limits_{}^{}(\cos^3(2t)+\cos^2(2t)+\cos(2t)+1)dt\]
Dido525
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I don't know what to do after :( .
Dido525
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The 4th step should be a dt.
anonymous
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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
Dido525
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Is there no other way?
Dido525
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I mean I did get a little further.
Dido525
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\[\frac{ -3 }{ 8 }\int\limits\limits_{}^{}(\cos^2(2t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]
anonymous
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i would start with
\[\int\cos^4(x)dx-\int\cos^6(x)dx\] and then look in the back of the book
Dido525
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\[\frac{ -3 }{ 8 }\int\limits\limits\limits_{}^{}((1-\sin^2(t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]
Dido525
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Where did that come from?
anonymous
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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
Dido525
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Yeah, I know. I turn them into the half angles, which I did.
Dido525
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That stupid cos^2(2t) is getting in the way.
anonymous
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\(\sin^2(x)=1-\cos^2(x)\) 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
Dido525
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Using another half angle won't help because they are multiplied together which just makes it squared afain,
Dido525
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again*
anonymous
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why i find this topic rather dull. almost everything you need is printed on the back cover of your text
Dido525
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I know. I have to do it though.
anonymous
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then i really recommend looking up the formula for \(\int\cos^n(x)dx\)
Dido525
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So I get:\[\frac{ -3 }{ 8 }\int\limits_{}^{}(1-\sin^2(2t))\cos(2t)+(1-\sin^2(2t)+\cos(2t)+1)dt\]
Dido525
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Hmm good point.
Dido525
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Sec.
Dido525
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Can I ask a dumb question?
Dido525
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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
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sure so i can seem dumber if i cannot answer it
anonymous
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yes (whew)
anonymous
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ignore the annoying 3, put it in at the end
Dido525
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Hehe. Thanks. This should be easy now.
anonymous
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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 \(n=6\)
Dido525
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Ohh that's MASSIVELY convenient.
anonymous
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it is going to be a pain, but much less of a pain than what you were doing
Dido525
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Why 4 and 6?
Dido525
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I have 3 and 2.
anonymous
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\[\sin^2(x)\cos^4(x)=(1-\cos^2(x))\cos^4(x)=\cos^4(x)-\cos^6(x)\]
anonymous
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so 4 and 6
Dido525
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Right.
Dido525
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Thanks!
anonymous
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have fun!
anonymous
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yw