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Question

Question: How do you integrate \[\csc x\]?...

How do you integrate cscx\csc x?

Explanation

Solution

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,
1xdx=lnx+C\int {\dfrac{1}{x}dx = \ln \left| x \right| + C}

Complete step by step answer:
We have to calculate the integral of cscx\csc x.
I=csc(x)dxI = \int {\csc \left( x \right)dx}
Now, let us multiply and divide IIwith cscx+cotx\csc x + \cot x, we have,
I=cscxcscx+cotxcscx+cotxdxI = \int {\csc x\dfrac{{\csc x + \cot x}}{{\csc x + \cot x}}} dx
or I=csc2x+cscxcotxcscx+cotxdxI = \int {\dfrac{{{{\csc }^2}x + \csc x\cot x}}{{\csc x + \cot x}}dx}
Then, we perform a substitution,
Let u=cscx+cotxu = \csc x + \cot x, then
dudx=csc2xcscxcotx=(csc2x+cscxcotx)\dfrac{{du}}{{dx}} = - {\csc ^2}x - \csc x\cot x = - \left( {{{\csc }^2}x + \csc x\cot x} \right)
Hence,
I=1uduI = \int { - \dfrac{1}{u}du}
I=lnu+CI = - \ln \left| u \right| + C

Hence, the integration of cscx\csc x is lncscx+cotx+C - \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,
ba some expression\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.