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EscherichiaRinku
 4 years ago
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 >.<
EscherichiaRinku
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my 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
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0have you integrated this?

EscherichiaRinku
 4 years ago
Best ResponseYou've already chosen the best response.0my 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
 4 years ago
Best ResponseYou've already chosen the best response.0is this complete and full and final result?

EscherichiaRinku
 4 years ago
Best ResponseYou've already chosen the best response.0The 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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