Integration

[zinc dune]

Integral of this:

[misty pumice]

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[fresh trellis]

Use difference of squares on the numerator several times.
In the end, the whole thing should become a function of just sin(2x) and cos(2x).

[zinc dune]

i got -cos2x(sin^4 x + cos^4 x)

[fresh trellis]

Hm... sin(x)^4 + cos(x)^4 should also disappear.

[fresh trellis]

Maybe you meant -cos(2x)(sin(x)^2 + cos(x)^2), rather?

If so, then that's just -cos(2x), which is the correct simplified form.

[zinc dune]

how so

[fresh trellis]

Can you show what you did?

[zinc dune]

$\sin^8 x - \cos^8 x = (\sin^4x + cos^4x)(\sin^4x - cos^4x)$

[pine briar]

Invisible

[zinc dune]

only the second term can be simplified further

[fresh trellis]

No, not just the second.

Hint:
x^4 + y^4 = (x^4 + 2x^2 y^2 + y^4) - 2x^2 y^2

[zinc dune]

ohh right

yes i got it

ill keep this thread open for now

[fresh trellis]

Sure!

[zinc dune]

im solving more, ill ask here if i have more doubts

[zinc dune]

integral of this?

[fresh trellis]

Hm...
x^8 + 4 = (x^8 + 4x^4 + 4) - 4x^4 = ...
You can continue from here.

[zinc dune]

i figured

[fresh trellis]

Well, pretty standard: complete the square, then use a reduction formula.

[zinc dune]

figured