Definite Integration, keep getting wrong answer. 
Correct to 4 significant figures
  $$\int_{1}^{2}{\csc^24tdt}$$

Done this multiple times now and can't seem to get the answer at the back of the book.
Here's my attempt:
$$\int_{1}^{2}{\csc^24tdt}$$
$$-\dfrac{1}{4}\left[\cot 4(2)-\cot 4(1)\right]_{1}^{2}$$
$$-\dfrac{1}{4}\left[\frac{1}{\tan 8}-\frac{1}{\tan4}\right]$$
$$-\dfrac{1}{4}\left[-0.1471-0.8637\right]$$
$$-\dfrac{1}{4}\left[-1.0108\right]$$
$$0.2527$$
Back of the book has a similar answer which is $0.2572$
Where am I going wrong?
 A: The book is wrong. Plain and simple. They transposed the last two digits, which is a simple typo. Originally, I was going to write the following, and I include it here for numerical reference.
You need to do all the rounding right at the end. Since we have calculators today, which usually carry 12-16 digits of precision, round-off error is the least of our worries. If you carry out the calculations without intermediate rounding, you get an answer of: $0.2527$.
As an aside, you cannot evaluate the integral directly, since it passes through a discontinuity at $\frac{\pi}{2}$, as shown below:

A: The antiderivative is perfectly correct $$\int{\csc^2(4t)\,dt}=-\frac{1}{4} \cot (4 t)$$ but, as already said in comments and answers, $\cot (4 t)$ shows a discontinuity at $t=\frac \pi 2$ and this value is just between the bounds. So, let us consider $$I=\int_1^{\frac \pi 2-\epsilon}{\csc^2(4t)\,dt}=\frac{1}{4} (\cot (4 \epsilon )+\cot (4))$$ $$J=\int_{\frac \pi 2+\epsilon}^2{\csc^2(4t)\,dt}=\frac{1}{4} (\cot (4 \epsilon )-\cot (8))$$ $$I+J=\frac{1}{2} \cot (4 \epsilon )+\frac{1}{4}\csc (8)$$ The last term $\frac{1}{4}\csc (8)\approx 0.252689$ corresponds to the answer of the book but the problem is really with the first terms $\frac{1}{2} \cot (4 \epsilon )$. Using Taylor series built at $\epsilon=0$, $$\frac{1}{2} \cot (4 \epsilon )=\frac{1}{8 \epsilon }-\frac{2 \epsilon }{3}+O\left(\epsilon ^3\right)$$ which shows the problem perfectly illustrated by FundThmCalculus's answer.
