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- EscherichiaRinku

Need some help integrating this and finding the coefficients (one given here as c and other comes from integration).
\[y = \frac{c}{2}\int\limits \frac{1}{1-\sin^3(3x)}dx\]
with following boundary constraints:
\[y(-\frac{\pi}{12})=\frac{1}{2},\ y(\frac{\pi}{12})=\frac{3}{2}\]
I can't believe this is an exam exercise, it's takes too long to solve it than it's worth the points >.<

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- EscherichiaRinku

- katieb

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- anonymous

my guess is that you cannot find a nice closed form for this integral, but i will be quiet and let a smarter person try

- EscherichiaRinku

I tried both substituting the denominator or substitute sin(3x), but both ways that one resulted in a way too long formula x_x;

- Shayaan_Mustafa

have you integrated this?

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- EscherichiaRinku

my simpler result our of two looks like this:
\[-\frac{9c}{4}(\sin^2(3x)cox(3x))+\frac{1}{4}(\ln(1-\sin^3(3x))+\frac{1}{2}K\]
and its quite a mess to find the constants. I wonder if this integration result is even correct o.0

- Shayaan_Mustafa

is this complete and full and final result?

- EscherichiaRinku

The integration is complete (not necesarily without errors) and this still needs the constants c and K, which are also quite a pain to find :(

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