Does this vanish? $\left[\left((12\cos^2t+K^2\sin^2t)^{(S+3)/2}-(12\cos^2t)^{(S+3)/2}\right)(-\cot t)\right]^\pi_0$ I have a problem that has kept me awake for several nights XD
If I integrate this (where $K$ and $S$ are constants;$K$ is small and $S$ can take values between 0 and 1):
$$\int_{0}^{\pi}\frac{1}{\sin^2t}\left[\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right]\mathrm{d}t$$
by parts, I get this:
$$=\left[ \left(\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right)(-\cot t)\right]^\pi_0-\int_{0}^{\pi}[...]$$
My question is: does the term
$$\left[ \left(\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right)(-\cot t)\right]^\pi_0$$ vanish when the limits of integration are applied? Someone$^*$ says yes, but why?
Thanks in advance!
Victor
$^*$ here is a statement about it from a French-speaking mathematician:

It reads (more or less)

One can integrate by parts $$\int\frac{dt}{\sin^2t}[\cdots]=$$
This expression simplifies.

*

*Indeed, the terms in brackets in the first line cancel, due to $\sin^2t/\sin t$.


Edit: Can be used L'Hospital's rule to check that?
 A: I've worked this problem out and found, it's the result of an improper integral, where L'Hospital's rule can be applied. That is, if $\mathrm{cot}\;t=\frac{\mathrm{cos}\;t}{\mathrm{sin}\;t}$, we'll get:
$$\underset{t\to \pi }{\mathrm{lim}} \left\lbrack {\left(12\;{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12\;{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(-\frac{\mathrm{cos}\;t}{\mathrm{sin}\;t}\right)$$
$$=\underset{t\to \pi }{\mathrm{lim}} \frac{\left\lbrack {\left(12\;{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12\;{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(-\mathrm{cos}\;t\right)}{\mathrm{sin}\;t}$$
$$\overset{L\prime \mathrm{Hôpital}}{=} \underset{t\to \pi }{\mathrm{lim}} \frac{\left(\frac{s+3}{2}\right)\left\lbrack {\left(12{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+1}{2}} \left(K^2 -12\right)+12{\left(12{\mathrm{cos}}^2 \;t\right)}^{\frac{s+1}{2}} \right\rbrack \left(\mathrm{sin}\;2t\right)\left(-\mathrm{cos}\;t\right)+\cdots }{\mathrm{cos}\;t}$$
$$\cdots \frac{\left\lbrack {\left(12{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(\mathrm{sin}\;t\right)}{\mathrm{cos}\;t}$$
Now, if $\mathrm{sin}\;2\pi =0$, we'll get:
$$=\frac{\left(\frac{s+3}{2}\right)\left\lbrack {\cdots} \right\rbrack \left(0\right)\left(1\right)+\left\lbrack {\cdots} \right\rbrack \left(0\right)}{-1}=0$$
Similarly, if $\mathrm{sin}\;0 =0$, we'll get:
$$=\frac{\left(\frac{s+3}{2}\right)\left\lbrack {\cdots} \right\rbrack \left(0\right)\left(-1\right)+\left\lbrack {\cdots} \right\rbrack \left(0\right)}{1}=0$$
:-)
P.S. In case that somebody needs it, I can explain how the derivative was made.
