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 .

Question

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 >.<

anonymous · Answer

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 · Answer

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 · Answer

have you integrated this?

escherichiarinku · Answer

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 · Answer

is this complete and full and final result?

escherichiarinku · Answer

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 :(