Question

Evaluate the intergal

Answer 1

\[\int\limits_{}^{}3\sin^2(t)\cos^4(t)dt\]

Answer 2

My work:

\[3\int\limits_{}^{}\sin^2(x)\cos^4(t)dt\]

\[3\int\limits_{}^{}\sin^2(t)\cos^2(t)\cos^2(t)dt\]

\[3\int\limits_{}^{}\frac{ 1 }{ 2 }(1-\cos(2t))(\frac{ 1 }{ 2 }(1+\cos(2t)))^2dt\]

\[\frac{ 3 }{ 8 }\int\limits_{}^{}(1-\cos(2t))((1+\cos(2t)))^2dx\]

\[\frac{ -3 }{ 8 }\int\limits_{}^{}((\cos^3(2t)+\cos^2(2t)+\cos(2t)+1) dt\]

\[\frac{ -3 }{ 8 }\int\limits_{}^{}((cos^2(2t))(cos(2t))+\cos^2(2t)+\cos(2t)+1))dt\]

\[\frac{ -3 }{ 8 }\int\limits_{}^{}(\cos^3(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

Answer 3

I don't know what to do after :( .

Answer 4

The 4th step should be a dt.

Answer 5

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

Answer 6

Is there no other way?

Answer 7

I mean I did get a little further.

Answer 8

\[\frac{ -3 }{ 8 }\int\limits\limits_{}^{}(\cos^2(2t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

Answer 9

i would start with
\[\int\cos^4(x)dx-\int\cos^6(x)dx\] and then look in the back of the book

Answer 10

\[\frac{ -3 }{ 8 }\int\limits\limits\limits_{}^{}((1-\sin^2(t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

Answer 11

Where did that come from?

Answer 12

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

Answer 13

Yeah, I know. I turn them into the half angles, which I did.

Answer 14

That stupid cos^2(2t) is getting in the way.

Answer 15

\(\sin^2(x)=1-\cos^2(x)\) 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

Answer 16

Using another half angle won't help because they are multiplied together which just makes it squared afain,

Answer 17

again*

Answer 18

why i find this topic rather dull. almost everything you need is printed on the back cover of your text

Answer 19

I know. I have to do it though.

Answer 20

then i really recommend looking up the formula for \(\int\cos^n(x)dx\)

Answer 21

So I get:\[\frac{ -3 }{ 8 }\int\limits_{}^{}(1-\sin^2(2t))\cos(2t)+(1-\sin^2(2t)+\cos(2t)+1)dt\]

Answer 22

Hmm good point.

Answer 23

Sec.

Answer 24

Can I ask a dumb question?

Answer 25

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?

Answer 26

sure so i can seem dumber if i cannot answer it

Answer 27

yes (whew)

Answer 28

ignore the annoying 3, put it in at the end

Answer 29

Hehe. Thanks. This should be easy now.

Answer 30

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 \(n=6\)

Answer 31

Ohh that's MASSIVELY convenient.

Answer 32

it is going to be a pain, but much less of a pain than what you were doing

Answer 33

Why 4 and 6?

Answer 34

I have 3 and 2.

Answer 35

\[\sin^2(x)\cos^4(x)=(1-\cos^2(x))\cos^4(x)=\cos^4(x)-\cos^6(x)\]

Answer 36

so 4 and 6

Answer 37

Right.

Answer 38

Thanks!

Answer 39

have fun!

Answer 40

yw