How do you integrate \[\csc x\]?

Hint: In the given question, we have been given a trigonometric function. We have to integrate it. To do that, we multiply and divide the integral with an expression including two trigonometric functions – ‘cosecant’ and ‘cotangent’. Then we make substitutions and use the standard integration result of the logarithm to calculate the answer.

Formula Used:
We are going to use the formula of integration of the reciprocal of a variable, which is,
\[\int {\dfrac{1}{x}dx = \ln \left| x \right| + C} \]

Complete step by step answer:
We have to calculate the integral of \[\csc x\].
\[I = \int {\csc \left( x \right)dx} \]
Now, let us multiply and divide \[I\]with \[\csc x + \cot x\], we have,
\[I = \int {\csc x\dfrac{{\csc x + \cot x}}{{\csc x + \cot x}}} dx\]
or \[I = \int {\dfrac{{{{\csc }^2}x + \csc x\cot x}}{{\csc x + \cot x}}dx} \]
Then, we perform a substitution,
Let \[u = \csc x + \cot x\], then
\[\dfrac{{du}}{{dx}} = - {\csc ^2}x - \csc x\cot x = - \left( {{{\csc }^2}x + \csc x\cot x} \right)\]
Hence,
\[I = \int { - \dfrac{1}{u}du} \]
\[I = - \ln \left| u \right| + C\]

Hence, the integration of \[\csc x\] is \[ - \ln \left| {\csc x + \cot x} \right| + C\].

Additional Information:
If we have a definite integral, then we calculate its value by putting in the upper limit into the result, then putting in the lower limit into the result and then subtracting the two. A definite integral is the one which looks like,
\[\int\limits_b^a \text{ some expression} \].

Note:
In the given question, we had to find the integral of cosecant. We did that by multiplying and dividing the integral with a particular expression. Then we did the calculations, simplifications of the given expression. Then we used the standard result of a logarithm into the integral to find the answer.