Evaluate the intergal

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Evaluate the intergal

Mathematics
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\[\int\limits_{}^{}3\sin^2(t)\cos^4(t)dt\]
My work: \[3\int\limits_{}^{}\sin^2(x)\cos^4(t)dt\] \[3\int\limits_{}^{}\sin^2(t)\cos^2(t)\cos^2(t)dt\] \[3\int\limits_{}^{}\frac{ 1 }{ 2 }(1-\cos(2t))(\frac{ 1 }{ 2 }(1+\cos(2t)))^2dt\] \[\frac{ 3 }{ 8 }\int\limits_{}^{}(1-\cos(2t))((1+\cos(2t)))^2dx\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}((\cos^3(2t)+\cos^2(2t)+\cos(2t)+1) dt\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}((cos^2(2t))(cos(2t))+\cos^2(2t)+\cos(2t)+1))dt\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}(\cos^3(2t)+\cos^2(2t)+\cos(2t)+1)dt\]
I don't know what to do after :( .

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

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