How do I go about solving the integral of csc x?
So here is a solution provided by another user a couple of years ago and I've seen this solution before but I'm not clear on how or why?
What would make me think or tip me off that I should multiply it the way that is shown?
Also it looks like there is a u-sub but I don't understand how it gets to the 1/u phase.
It would be greatly appreciated if you could help me understand the solution better.
Thanks in advance!
$\endgroup$ 33 Answers
$\begingroup$Hoping that I am not of topic, there is a very simple way to compute this antiderivative : the tangent half-angle substitution $t=\tan(\frac x2)$.
So $$I=\int \csc (x)\,dx=\int \frac {dx}{\sin(x)}=\int \frac {dt}{t}=\log(t)=\log\Big(\tan\big(\frac x2\big)\Big)$$
$\endgroup$ 2 $\begingroup$Claude Leibovici's answer gives the easiest way to integrate this.
Another method that's longer than the one you refer to, but is maybe easier to motivate, is to write
$\displaystyle\int\csc x\;dx=\int\frac{1}{\sin x}\;dx=\int\frac{\sin x}{\sin^2x}\;dx=\int\frac{\sin x}{1-\cos^2 x}\;dx$.
Now let $u=\cos x, \;du=-\sin xdx$ to get $\displaystyle\int\frac{1}{u^2-1}\;du$, and then use partial fractions.
$\endgroup$ 3 $\begingroup$$u=\csc(x)-\cot(x)$, so $du = \csc^2(x)-\cot(x)\csc(x) dx$.
\begin{align*} \int\frac{\csc^{2}(x)-\csc(x)\cot(x)}{\csc(x)-\cot(x)}dx &= \\ \int\frac{\csc^{2}(x)-\csc(x)\cot(x)}{u}\frac{1}{\csc^2(x)-\cot(x)\csc(x)} du &= \\ \int\frac{1}{u} du \end{align*}
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