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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

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

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