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Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}3\sin^2(t)\cos^4(t)dt\]

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0My 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
 2 years ago
Best ResponseYou've already chosen the best response.0I don't know what to do after :( .

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0The 4th step should be a dt.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1look 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
 2 years ago
Best ResponseYou've already chosen the best response.0I mean I did get a little further.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 3 }{ 8 }\int\limits\limits_{}^{}(\cos^2(2t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i would start with \[\int\cos^4(x)dx\int\cos^6(x)dx\] and then look in the back of the book

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 3 }{ 8 }\int\limits\limits\limits_{}^{}((1\sin^2(t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0Where did that come from?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1there 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
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah, I know. I turn them into the half angles, which I did.

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0That stupid cos^2(2t) is getting in the way.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1\(\sin^2(x)=1\cos^2(x)\) 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
 2 years ago
Best ResponseYou've already chosen the best response.0Using another half angle won't help because they are multiplied together which just makes it squared afain,

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1why i find this topic rather dull. almost everything you need is printed on the back cover of your text

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0I know. I have to do it though.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1then i really recommend looking up the formula for \(\int\cos^n(x)dx\)

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0So I get:\[\frac{ 3 }{ 8 }\int\limits_{}^{}(1\sin^2(2t))\cos(2t)+(1\sin^2(2t)+\cos(2t)+1)dt\]

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0Can I ask a dumb question?

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0We have that constant out in front. If we break up the intergal into seperate sums do I distribute that constant to all the intergals?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1sure so i can seem dumber if i cannot answer it

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1ignore the annoying 3, put it in at the end

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0Hehe. Thanks. This should be easy now.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1here it is \[\int\cos^n(x)dx = \frac{1}{n}\cos(x)\sin(x)+\frac{n1}{n}\int \cos^{n2}(x)dx\] use with \(n=4\) and again with \(n=6\)

Dido525
 2 years ago
Best ResponseYou've already chosen the best response.0Ohh that's MASSIVELY convenient.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1it is going to be a pain, but much less of a pain than what you were doing

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1\[\sin^2(x)\cos^4(x)=(1\cos^2(x))\cos^4(x)=\cos^4(x)\cos^6(x)\]
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