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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?

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  • $\begingroup$ First of all, are you sure this can be integrated over that range? What happens when $t = \frac{\pi}{2}$? $\endgroup$ – Deepak Jun 30 '15 at 1:15
  • $\begingroup$ No idea, I'm just doing questions from the book, (John Bird, Engineering Mathematics) $\endgroup$ – Modrisco Jun 30 '15 at 1:16
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    $\begingroup$ wolframalpha.com/input/… symbolab.com/solver/calculus-calculator/… The area must be infinite, and not the answer you've received, due to the presence of $\frac{\pi}{2}$ between 1 and 2, where the curve tends towards infinity. $\endgroup$ – Kugelblitz Jun 30 '15 at 1:19
  • $\begingroup$ Apologies, I get what you mean now @Deepak I wonder how I was able to receive an answer even though the area must is infinite. $\endgroup$ – Modrisco Jun 30 '15 at 1:24
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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:

discontinuity marked in dashed red line

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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.

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