Integration of cosec x

Integration of cosec x can be found using various integration technique like integration by substitution, Integration by partial fraction along with trigonometric identities. The various formula for integration of cosec x are

I

\[
\int \csc(x) \, dx = \ln | \csc(x) – \cot(x) | + C
\]

II

\[
\int \csc(x) \, dx = -\ln | \csc(x) + \cot(x) | + C
\]

III

\[
\int \csc(x) \, dx = \ln | \tan \frac {x}{2} | + C
\]

IV

\[
\int \csc(x) \, dx = \frac {1}{2} \ln | \frac {\cos x -1}{\cos x + 1} | + C
\]

Lets check out the proof of each of these

Proof of I

Integration of cosec x can be solved by using a clever trick involving multiplying and dividing by $(\csc(x) – \cot(x))$. Here’s how it’s done:

\[
\int \csc(x) \, dx = \int \csc(x) \cdot \frac{\csc(x) – \cot(x)}{\csc(x) -\cot(x)} \, dx \\
= \int \frac{\csc^2(x) – \csc(x)\cot(x)}{\csc(x) – \cot(x)} \, dx.
\]

Now, taking u as
$ u= \csc(x) – \cot(x) $
$ du =(-\csc(x)\cot(x) + \csc^2(x)) dx $

Therefore integration becomes

\[
\int \csc(x) \, dx =\int \frac {1}{u} \, du,
\]

hence

\[
\int \csc(x) \, dx = \ln | \csc(x) – \cot(x) | + C
\]

where (C) is the constant of integration. This is the integral of $(\csc(x))$.

Proof of II

We can write the result of I as

\[
\int \csc(x) \, dx = \ln | \csc(x) – \cot(x) | + C = \ln | \frac {\csc^2(x) – \cot^2(x)}{\csc(x) + \cot(x) } | + C \\
=\ln | \frac {1}{\csc(x) + \cot(x) } | + C = \ln | (\csc(x) + \cot(x))^{-1} | + C \\
=- \ln | \csc(x) + \cot(x) | + C
\]

Proof of III

\[
\int \csc(x) \, dx =\int \frac {1}{\sin(x)} \, dx
=\int \frac {1}{2 \sin(x/2) \cos(x/2)} \, dx
\]

Dividing and Multiplying cos(x/2) in the denominator

\[
\int \csc(x) \, dx = \frac {1}{2}\int \frac {\sec^2(x/2)}{\tan(x/2)} \, dx
\]

Lets $u=\tan(x/2)$
$du= \frac{1}{2} \sec^2(x/2) dx$

Therefore

\[
\int \csc(x) \, dx = \int \frac {1}{u} \, du = ln |u| + C
\]

Hence

\[
\int \csc(x) \, dx = \ln | \tan \frac {x}{2} | + C
\]

Proof of IV

This can be proof by using integration by partial fractions

\[
\int \csc(x) \, dx =\int \frac {1}{\sin(x)} \, dx
=\int \frac {sin(x)}{\sin^2(x)} \, dx
=\int \frac {sin(x)}{ 1- \cos^2(x)} \, dx
\]

Lets $ u=\cos(x)$
$du= -\sin(x) dx$

Therefore

\[
\int \csc(x) \, dx =\int \frac {1}{u^2 -1} \, du
=\frac {1}{2} \int [\frac {1}{u-1} – \frac {1}{u+1} ]\, du
=\frac {1}{2} ln |\frac {u-1}{u+1}| + C
\]

Hence

\[
\int \csc(x) \, dx = \frac {1}{2} \ln | \frac {\cos x -1}{\cos x + 1} | + C
\]

Solved Examples

Question
Evaluate the definite integral $\int_{\pi/6}^{\pi/3} \csc(x) \, dx $.
Solution

Using the result from the above

\[
\int_{\pi/6}^{\pi/3} \csc(x) \, dx = \left[- \ln | \csc(x) + \cot(x) | \right]_{\pi/6}^{\pi/3} \\
= -\ln | \csc(\pi/3) + \cot(\pi/3) | + \ln | \csc(\pi/6) + \cot(\pi/6) | \\
= \ln | 2 |- \ln | 2 + \sqrt{3} | \\
= \ln(\frac {2}{2+ \sqrt 3}).
\]