Finding the Integral of csc(x)

In this lesson, we will see the steps involved with finding the integral of csc(x). We will find a solution, and then check our work using derivatives.
Steps to Solving the Problem
The problem of finding the integral of cscx involves integration and trigonometric functions. Integration is the process of finding the integral of something, and trigonometric functions are functions of angles. Both of these things come in very handy in engineering, construction, astronomy, medicine, and chemistry, as well as in many other areas. Our problem in particular is asking us to find the integral of cscx. There are three main steps in finding this integral.
The first step is definitely not obvious. We want to start out by multiplying and dividing cscx by cscx + cotx. As we go along with this problem, it will become clear as to why we want to do this.
Multiplying and Dividing by the Same Expression
integral cscx1
It is okay to do this because multiplying and dividing by the same expression is the same as multiplying by 1. Thus, it doesn’t change the value of cscx.
The next step is to simplify a bit. We multiply cscx through the numerator. We are also going to factor a negative out of the numerator and put it outside the integral. Again, we will see why we do this in Step 3.
integral cscx2
Now that we’ve done this, we have the integral of -(-csc2x – cscx cotx) / (cscx + cotx).
The third step is the step where it becomes clear why we’ve done Steps 1 and 2. We are going to be using substitution to find the integral. Substitution is basically the chain rule for derivatives in reverse.
Integral U substitution
integral cscx3
Notice that the denominator is cscx + cotx (from Step 2), and the derivative of this is -cscx cotx – csc2x dx, which is equal to – csc2x – cscx cotx dx. We see that this is our numerator. Thus, we are going to make the substitution u = cscx + cotx, so the derivative of u is du = -csc2x – cscx cotx dx. Now we can plug these into our integral from Step 2.
Substitution in Our Example
integral cscx8
We see that we now have the integral of -1/u du. This is a well-known integral, and it is equal to -ln |u| + C, where C is a constant.
Our last step is to put our value for u (cscx + cotx) back into our answer. This gives -ln |cscx + cotx| + C, where C is a constant.
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