# 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?

• First of all, are you sure this can be integrated over that range? What happens when $t = \frac{\pi}{2}$? – Deepak Jun 30 '15 at 1:15
• No idea, I'm just doing questions from the book, (John Bird, Engineering Mathematics) – Modrisco Jun 30 '15 at 1:16
• 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. – Kugelblitz Jun 30 '15 at 1:19
• 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. – Modrisco Jun 30 '15 at 1:24

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:
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.