I'm not sure trig identity to use. Would we use $1+\cot^2(x) = \csc^2(x)$? I know that we break down into $\csc^2(x)$ and $\csc(x)$. Could I get hints?
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Try integration by parts, with $u=\csc x$ and $dv=\csc^{2}x dx$, and then use your identity afterwards to solve for the integral. – user84413 Oct 16 '14 at 16:03
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Apologies! The approach I suggested does not work--I misremembered my trig derivatives. – Cameron Buie Oct 16 '14 at 16:36
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Use the integration by parts technique and try to derive and equation with $ \int{csc^3(x)dx}$ as your unknown parameter. Then, solve the equation.
$$\int{csc^3(x)}dx$$
$$ u = csc(x) \text{ } dv = csc^2(x)$$
$$ du = -cot(x)csc(x) \text{ } v = -cot(x) $$
$$\int{csc^3(x)}dx = -csc(x)cot(x) - \int{cot(x)cot(x)csc(x)dx}$$
$$ = -csc(x)cot(x) - \int{cot^2(x)csc(x)dx}$$
Now use $cot^2(x) = csc^2(x) - 1$
$$ = -csc(x)cot(x) - \int{(csc^2(x) - 1)csc(x)dx}$$
$$ = -csc(x)cot(x) - \int{csc^3(x)dx} +\int{csc(x)dx}$$
That is,
$$\int{csc^3(x)}dx = -csc(x)cot(x) - \int{csc^3(x)dx} +\int{csc(x)dx}$$
Then
$$\int{csc^3(x)}dx = \frac{1}{2}(-csc(x)cot(x) +\int{csc(x)dx})$$
And
$$\int{csc(x)dx} = log(tan(x/2)) $$
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Use the recurent formula $$\int\csc^n x dx=-\frac{1}{n-1}\cot x\csc^{n-2}x+\frac{n-2}{n-1}\int\csc^{n-2}xdx$$ for $n=3$ we have $$\int\csc^3 xdx=-\frac{1}{2}\cot x\csc x+\frac{1}{2}\int\csc x dx=-\frac{1}{2}\cot x\csc x+\frac{1}{2}\ln\vert\csc x- \cot x\vert+C$$
Madrit Zhaku
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