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Calculus: Tutoring Solution13 chapters | 111 lessons

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Lesson Transcript

Instructor:
*Laura Pennington*

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

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.

The problem of finding the integral of csc*x* 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 csc*x*. 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 csc*x* by csc*x* + cot*x*. As we go along with this problem, it will become clear as to why we want to do this.

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 csc*x*.

The next step is to simplify a bit. We multiply csc*x* 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.

Now that we've done this, we have the integral of -(-csc2*x* - csc*x* cot*x*) / (csc*x* + cot*x*).

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.

Notice that the denominator is csc*x* + cot*x* (from Step 2), and the derivative of this is -csc*x* cot*x* - csc2*x* *dx*, which is equal to - csc2*x* - csc*x* cot*x* *dx*. We see that this is our numerator. Thus, we are going to make the substitution *u* = csc*x* + cot*x*, so the derivative of *u* is *du* = -csc2*x* - csc*x* cot*x* *dx*. Now we can plug these into our integral from Step 2.

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* (csc*x* + cot*x*) back into our answer. This gives -ln |csc*x* + cot*x*| + *C*, where *C* is a constant.

We've found that the integral of csc*x* is -ln |csc*x* + cot*x*| + *C*.

As we said, integration is the finding of the integral (or anti-derivative) of something. Integration basically undoes finding the derivative of an expression. That is, if *a* is the integral of *b*, then the derivative of *a* is *b*. We can use this fact to check our answer. We found that -ln |csc*x* + cot*x*| + *C* is the integral of csc*x*. Based on our fact, it should be the case that the derivative of -ln |csc*x* + cot*x*| + *C* is csc*x*.

We see that to check our answer, we just have to find the derivative of -ln |csc*x* + cot*x*| + *C*. For starters, we can drop the *C*, because when we take the derivative of a constant, we get zero, so that will fall off. To find the derivative of -ln |csc*x* + cot*x*|, we use the following information:

First we use the chain rule, where *f*(*x*) = -ln |*x*|, and *g*(*x*) = csc*x* + cot*x*.

Then, note that the derivative of a sum of functions is the sum of the derivatives of each of those functions. That is, (*f*(*x*) + *g*(*x*)) = *f*(*x*) + *g*(*x*).

The derivative of ln |*x*| = 1/*x*. The derivative of csc*x* is -csc*x* cot*x*. The derivative of cot*x* is -csc2*x*.

Plugging our *f*(*x*) and *g*(*x*) into the chain rule, we get that the derivative of -ln |csc*x* + cot*x*| is (-1 / csc*x* + cot*x*) * (-csc*x* cot*x* - csc2*x*).

We see that the derivative of -ln |csc*x* + cot*x*| + *C* is csc*x*, and this tells us that our answer is correct. Therefore, the integral of csc*x* is -ln |csc*x* + cot*x*| + *C*.

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13 in chapter 10 of the course:

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Calculus: Tutoring Solution13 chapters | 111 lessons

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