Question

Integrate:
cos^4 x - sin^4 x . dx

Answer 1

\[y = \int\limits (\cos^4x-\sin^4x).dx\]

Answer 2

@zepdrix

Answer 3

Hmm I just worked it out on paper... I must be making a mistake somewhere, there's no way it can really be that simple...

I'll show you what I came up with, maybe you can see my mistake :3 if there is one.

Answer 4

I got till:
\[\frac 14 \int\limits 4 \cos 2x + 2 \cos^22x\]But it seems wrong.

Answer 5

\[\large \int\limits \cos^4x-\sin^4x dx \quad \rightarrow \quad \int\limits (\cos^2x)^2-\sin^4x dx\]
\[\large \rightarrow \quad \int\limits(1-\sin^2x)^2-\sin^4x dx\]
\[\large \rightarrow \quad \int\limits (1-2\sin^2x+\sin^4x)-\sin^4x dx\]
\[\large \rightarrow \quad \int\limits 1-2\sin^2x dx\]And from here we can use the Double Angle Formula for Sine. Did I expand that square correctly though? :o hmm

Answer 6

This is how i was doing.. Ended up nowhere. -_-

Answer 7

Actually I think you were headed for the same answer :) I see a silly little mistake near the end I think.. one sec, checking.

Answer 8

You should have ended up with this, the square terms should've cancelled out.
\[\large \frac{1}{4}\int\limits 4\cos(2x) dx\]

Answer 9

Which is equivalent to what I have above after you apply the double angle formula.

Answer 10

Just missed a minus on the final square term I think :D

Answer 11

Ooooooooooooooooo! Nice catch!

Answer 12

Thanks man! :)

Answer 13

Hmm fun problem :D They test you on your ability to mess around with trig identities, then you get a nice clean answer in the end!

Answer 14

Truee!!
So final answer would be?

Answer 15

\[\large \int\limits \cos(2x)dx \quad = \quad \frac{1}{2}\sin(2x)+c\]And if you really wanted to (Since we started with an argument of (x) ), you could apply the Double Angle Formula one last time, giving you\[\large \frac{1}{2}(2sinx\cdot cosx)\quad =\quad \sin x\cdot \cos x\]

Answer 16

Oh, plus C :D whatev lol

Answer 17

Oh damn it. My bad. lol.
i got till 1/2 sin(2x) + C

Thanks again.