I found this question in an old real analysis text book, (so old the cover had come off) I graphed it on wolfram alpha, and it looks like an almost odd function. I think the integral evaluates to zero. Would anyone care help prove? (assuming I'm right)


1 Answer 1


$\sin 4x = 2 \cos 2x \sin 2x = 4 (2 \sin^2 x -1 ) \sin x \cos x $

Substitute $t = \sin x$

  • $\begingroup$ dang that was easy, I've been getting lazy... Thank you so much! $\endgroup$
    – ptr64
    Sep 30, 2015 at 22:51

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