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Alright so I've got the question:

$\int2\sin^2(x)\cos^2(x)dx$

And in class I learned:

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

$\cos^2(x) = ((1+\cos(2x))/2)$

So when I substitute I get:

$\int2((1-\cos(2x))/2)((1+\cos(2x))/2)dx$

But according to the almighty wolfram, those two aren't the same integral. What did I do wrong? And furthermore, how to I proceed?

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    $\begingroup$ You should use LaTeX for the math parts of your questions. It makes people answer faster. $\endgroup$ – user12205 Jun 23 '11 at 0:46
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    $\begingroup$ $\int 2 sin^{2}x cos^{2}xdx$? do you know $2 sin x cos x = sin 2x $? $\endgroup$ – newbie Jun 23 '11 at 0:49
  • $\begingroup$ I'm having trouble finding a LaTeX tutorial online. Could you point me to one? $\endgroup$ – InBetween Jun 23 '11 at 0:53
  • $\begingroup$ Ah nvm, someone edited my post and I looked at the edits, pretty straight forward. $\endgroup$ – InBetween Jun 23 '11 at 0:55
  • $\begingroup$ @InBetween: tobi.oetiker.ch/lshort/lshort.pdf $\endgroup$ – user12205 Jun 23 '11 at 1:00
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When I put each of them into WolframAlpha, I got the same thing:

original integral

second integral

However, if I were evaluating your original integral by hand, I'd probably have used the identity $\sin 2x=2\sin x\cos x$ first (any time I see $\sin x\cos x$, that identity comes to mind).

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  • $\begingroup$ Ah, I was missing a set a parenthesis. Dumb mistake. Thanks for the help and the other identity! $\endgroup$ – InBetween Jun 23 '11 at 1:22
  • $\begingroup$ @InBetween: Wolfram Alpha gives you a fairly nicely formatted display of what it "thinks" you mean. It pays to look carefully at that display. $\endgroup$ – André Nicolas Jun 23 '11 at 1:32

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