I am trying to understand the solution to an arc length question.
Integrating: $$ =\int_{\frac{\pi}{9}}^{\frac{5 \pi}{18}} \csc (3 x) d x $$
The workbook has: $$ =\left.\frac{-1}{3} \ln (\csc (3 x)+\cot (3 x))\right|_{\frac{\pi}{9}} ^{\frac{5 \pi}{18}} $$
where does the negative come $\frac{-1}{3}$ from?
I thought the integral would have been: $$ =\left.\frac{1}{3} \ln (\csc (3 x)+\cot (3 x))\right|_{\frac{\pi}{9}} ^{\frac{5 \pi}{18}} $$
What am I doing wrong?
On
Well, to show where the solution comes from we can write:
$$\csc x=\csc x\cdot\frac{\cot x+\csc x}{\cot x+\csc x}=\frac{\csc x\cot x+\csc^2 x}{\cot x+\csc x}\tag1$$
Now, when we integrate:
$$\int\csc x\space\text{d}x=\int\frac{\csc x\cot x+\csc^2 x}{\cot x+\csc x}\space\text{d}x\tag2$$
Subsitute $\text{u}=\cot x+\csc x$, so:
$$\int\csc x\space\text{d}x=\int-\frac{1}{\text{s}}\space\text{ds}=\text{C}-\ln\left|\text{s}\right|\tag3$$
Note:
$$\int \csc(x)dx = -\log |\csc(x)+\cot(x)|+C$$
So, the minus sign is there.