Question Video: Integration Involving the Reciprocal Trigonometric Functions Mathematics

Determine ∫csc 2π‘₯(57 sinΒ² 2π‘₯ + 33 cot 2π‘₯) dπ‘₯.

02:57

Video Transcript

Determine the integral of the csc of two π‘₯ multiplied by 57 sin squared of two π‘₯ plus 33 cot of two π‘₯ with respect to π‘₯.

In this question, we’re asked to evaluate the integral of a trigonometric function. And this is a very difficult-looking function to integrate. So let’s start by simplifying our integrand. First, we can notice we’re multiplying by the csc of two π‘₯, and we know this is equivalent to one divided by the sin of two π‘₯. When we distribute one divided by the sin of two π‘₯ over our parentheses, we’ll divide the first term inside the parentheses by the sin of two π‘₯ and the second term by this. However, we’ll write the second term as being multiplied by the csc of two π‘₯. Simplifying the first term, we get the integral of 57 times the sin of two π‘₯ plus 33 csc of two π‘₯ multiplied by the cot of two π‘₯ with respect to π‘₯.

And now, each of the two terms in our integrand are in standard forms which we know how to integrate by using our integral rules. So we can just integrate each term separately. Let’s start with the first term. We can recall for any real constant π‘Ž not equal to zero, the integral of the sin of π‘Žπ‘₯ with respect to π‘₯ is equal to negative one over π‘Ž multiplied by the cos of π‘Žπ‘₯ plus the constant of integration 𝑐. In our case, the value of π‘Ž is two. So we get negative 57 over two multiplied by the cos of two π‘₯. And remember, since we’re evaluating the indefinite integral of two terms, we can just add our constant of integration at the end.

Next, we need to evaluate the integral of 33 csc of two π‘₯ times the cot of two π‘₯. We can do this in a few different ways. For example, we know the derivative of the csc of π‘₯ with respect to π‘₯ is negative the csc of π‘₯ times the cot of π‘₯. This gives us a standard integral result. The integral of the csc of π‘₯ times the cot of π‘₯ with respect to π‘₯ is negative the csc of π‘₯ plus 𝑐. We can then use a 𝑒-substitution to rewrite our integrand in this form and then apply this rule. However, we can use the same process to determine a general rule in terms of any multiple argument.

We have for any real constants π‘˜ and π‘Ž, where π‘Ž is not equal to zero, the integral of π‘˜ times the csc of π‘Žπ‘₯ multiplied by the cot of π‘Žπ‘₯ with respect to π‘₯ is negative π‘˜ over π‘Ž times the csc of π‘Žπ‘₯ plus the constant of integration 𝑐. And it’s much easier to just apply this result. We have π‘˜ is 33 and π‘Ž is equal to two. This gives us negative 57 over two times the cos of two π‘₯ minus 33 over two times the csc of two π‘₯ plus 𝑐, which is our final answer. Therefore, by simplifying our integrand and applying our integral results, we were able to show the integral of the csc of two π‘₯ times 57 sin squared of two π‘₯ plus 33 cot of two π‘₯ with respect to π‘₯ is negative 57 over two cos of two π‘₯ minus 33 over two csc of two π‘₯ plus 𝑐.

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