C4 Integration Q Watch

Keoje
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#1
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#1
Integrate \frac1{sin^2 x cos^2 x}

From previous question I found that {sin^2 x cos^2 x} is the same as  \frac{1 - cos4x} {8}

Can someone give me a hint as to how to start of?

Thanks in advance.
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Winter
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#2
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sin2x= 2sinxcosx
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Swayum
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#3
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I wouldn't use (1-cos4x)/8 (I don't even know where that's coming from).

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

Can you see how to rewrite that as a single trig function?
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Keoje
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#4
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It comes from the Cos2x double angle formulae. It was for a prev. q neways.

And I can't see the single trig function =/
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Keoje
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#5
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double angle formulae (sin2x)/2 ?
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Swayum
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#6
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Yeah
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Keoje
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#7
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So far I've got...

1/(sin^2 x cos^2 x)

1/(sinxcosx)^2

1/((sin^2 2x)/4) (using sin2x formulae)

4/sin^2 2x (using cos2x: sin^2 2x = (1-cos4x)/2 )

8/(1-cos4x)

and I hit a wall here.

Help please
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Swayum
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#8
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From where you've got 4/sin^2 2x, don't use the cos2x identity.

4sin^2 2x = 4cosec^2 2x

Can you see how to integrate that? If not (and actually think about it before clicking the thing below)

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Keoje
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#9
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I got carried away using trig functions

Thanks a lot you two - I see how it works now. Will rep you both
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jenny0502
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#10
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ok.. I'd say forget the (1-cos4x)/2
You can't intergrate the function from there with C4 knowledge...

If you know (sinxcosx)^2 = ((sin2x)^2)/4

then you know that 1/(sinxcosx)^2 is 4/((sin2x)^2)

i.e int 4(sin2x)^-2 ?

I'm not sure why they made the (1-cos4x)/2 part of the first bit of the question.. it doesnt really seem to follow.. I could be wrong though
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jenny0502
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#11
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ahh.. going to make tea made me a little late there.. apologies :-)
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Keoje
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#12
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(1-cos4x)/2 I got when I went trig function crazy. :p:
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