8
$\begingroup$

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

$\endgroup$
19
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – Mathemagician1234 Mar 18 '13 at 17:28
10
$\begingroup$

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 $$\frac{dx}{\sin x}=\frac{dt}{t}=\ln(|t|)+C.$$

$\endgroup$
8
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.