Question

\[\ \int_{}^{} \cos^4 x -\sin^4 x ~~dx\]
i'm supposed to use this identity along with u-sub
\[\ \cos 2x =cos^2 x- sin^2 x \]

Answer 1

er one sec
latex messed up

Answer 2

nvm shows now

Answer 3

rewrite the function in terms of cos^2x-sin^2x

Answer 4

\[\cos ^{2}x-\sin ^{2}x\]

Answer 5

wouldn't work would it?
since it's a -sin^2x
\[\int (cos^2x - sin^2x)^2~~dx\]

Answer 6

or is the negative sign just not important

Answer 7

you have to add
\[\cos ^{2}xsin ^{2}x\]

Answer 8

I'm gonna let someone else take this one

Answer 9

I have an idea : \((cos^2)-(sin^2)^2 =(cos^2+sin^2)(cos^2-sin^2) = cos^2-sin^2= cos(2x)\)

Answer 10

the first term is \((cos^2)^2\)

Answer 11

ah

Answer 12

I thought of that as well

Answer 13

\[\ \int \cos^2 2x~~ dx \]
it ends up like this right?

Answer 14

nope, just cos (2x)

Answer 15

i mean
since
\[\ cos^2 x - sin^2x = cos 2x \]
then
\[\ cos^4x - sin^4x = cos^2x\]

Answer 16

I mean \(\ \int \cos( 2x)~~ dx\)

Answer 17

oops typo
\[\ cos^4 x - sin^4 x = cos^2 2x\]

Answer 18

does that work?

Answer 19

hahahah, how???

Answer 20

Ok, let me write the whole thing out

Answer 21

kk

Answer 22

Factor the original integrand as a difference of squares. One factor is = 1 (Trig ident) and the other is cos(2x)

Answer 23

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

Answer 24

like \(a^4 -b^4 = (a^2)^2 -(b^2)^2=(a^2+b^2)(a^2-b^2)\)

Answer 25

your a is cos, your b is sin, hence the first term a^2 +b^2 =1

Answer 26

Yes

Answer 27

Oh...
it makes sense now lol
thnx

Answer 28

hihihi.... students have that phase, me too, hihihi