# What is $\int\frac{dx}{\sin x}$?

I'm looking for the antiderivatives of $1/\sin x$. Is there even a closed form of the antiderivatives? Thanks in advance.

Hint: Write this as $$\int \frac{\sin (x)}{\sin^2 (x)} dx=\int \frac{\sin (x)}{1-\cos^2(x)} dx.$$ Now let $u=\cos(x)$, and use the fact that $$\frac{1}{1-u^2}=\frac{1}{2(1+u)}+\frac{1}{2(1-u)}.$$

Added: I want to give credit to my friend Fernando who taught me this approach.

• GEEZ! WHY DIDN'T ANYONE SHOW ME THIS TRICK WHEN I WAS A FRESHMAN CALCULUS STUDENT?!? Oh,I am so totally stealing this for my students,thanks.......lol – Mathemagician1234 Mar 18 '13 at 17:28

Let $R(u,v)$ be a rational function of $u$ and $v$, that is, a ratio of polynomials. Let $f(x)=R(\sin x, \cos x)$. Then there is a universal method for calculating $\int f(x)\,dx$, namely the Weierstrass substitution $t=\tan(x/2)$.

This substitution reduces the problem of integrating $f(x)$ to the problem of integrating a rational function of $t$. This can then be handled by the method of partial fractions.

Universal does not mean universally most efficient. We do not suggest the Weierstrass substitution for integrating $\cos x$!

Let's see what happens here. Let $t=\tan(x/2)$. It is standard (see the link above) that $\sin x=\frac{2t}{1+t^2}$ and $dx=\frac{2\,dt}{1+t^2}$. It follows that $$\frac{dx}{\sin x}=\frac{dt}{t}=\ln(|t|)+C.$$

$$\int\csc x\ dx=\int\frac{\csc x(\csc x-\cot x)}{\csc x-\cot x}\ dx=\int\frac{-\csc x\cot x+\csc^2 x}{\csc x-\cot x}\ dx=\ln |\csc x-\cot x|+c$$

As $\sin 2y=\frac{2\tan y}{1+\tan^2y}=\frac{2\tan y}{\sec^2y}$

$$\int\frac{dx}{\sin x}=\int \frac{(\sec^2\frac x2) dx}{2\tan \frac x2}=\ln\left| \tan \frac x2\right|+C$$ putting $\tan \frac x2=z\implies \frac12 \sec^2\frac x2 dx=dz$ where $C$ is the arbitrary constant for indefinite integral