WEBVTT
00:00:02.900 --> 00:00:14.020
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₯.
00:00:14.240 --> 00:00:18.480
In this question, weβre asked to evaluate the integral of a trigonometric function.
00:00:18.620 --> 00:00:21.300
And this is a very difficult-looking function to integrate.
00:00:21.330 --> 00:00:24.040
So letβs start by simplifying our integrand.
00:00:24.310 --> 00:00:32.190
First, we can notice weβre multiplying by the csc of two π₯, and we know this is equivalent to one divided by the sin of two π₯.
00:00:32.390 --> 00:00:40.880
When we distribute one divided by the sin of two π₯ over our parentheses, weβll divide the first term inside the parentheses by the sin of two π₯ and the second term by this.
00:00:40.910 --> 00:00:45.250
However, weβll write the second term as being multiplied by the csc of two π₯.
00:00:45.420 --> 00:00:55.800
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₯.
00:00:56.010 --> 00:01:02.370
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.
00:01:02.650 --> 00:01:04.980
So we can just integrate each term separately.
00:01:05.010 --> 00:01:06.360
Letβs start with the first term.
00:01:06.510 --> 00:01:18.800
We can recall for any real constant π not equal to zero, the integral of the sin of ππ₯ with respect to π₯ is equal to negative one over π multiplied by the cos of ππ₯ plus the constant of integration π.
00:01:19.070 --> 00:01:21.250
In our case, the value of π is two.
00:01:21.250 --> 00:01:25.630
So we get negative 57 over two multiplied by the cos of two π₯.
00:01:25.660 --> 00:01:32.030
And remember, since weβre evaluating the indefinite integral of two terms, we can just add our constant of integration at the end.
00:01:32.400 --> 00:01:38.490
Next, we need to evaluate the integral of 33 csc of two π₯ times the cot of two π₯.
00:01:38.680 --> 00:01:40.350
We can do this in a few different ways.
00:01:40.350 --> 00:01:47.440
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₯.
00:01:47.750 --> 00:01:49.490
This gives us a standard integral result.
00:01:49.490 --> 00:01:55.840
The integral of the csc of π₯ times the cot of π₯ with respect to π₯ is negative the csc of π₯ plus π.
00:01:56.190 --> 00:02:00.800
We can then use a π’-substitution to rewrite our integrand in this form and then apply this rule.
00:02:01.130 --> 00:02:06.300
However, we can use the same process to determine a general rule in terms of any multiple argument.
00:02:06.560 --> 00:02:21.250
We have for any real constants π and π, where π is not equal to zero, the integral of π times the csc of ππ₯ multiplied by the cot of ππ₯ with respect to π₯ is negative π over π times the csc of ππ₯ plus the constant of integration π.
00:02:21.530 --> 00:02:23.670
And itβs much easier to just apply this result.
00:02:23.670 --> 00:02:26.670
We have π is 33 and π is equal to two.
00:02:26.990 --> 00:02:35.380
This gives us negative 57 over two times the cos of two π₯ minus 33 over two times the csc of two π₯ plus π, which is our final answer.
00:02:35.710 --> 00:02:54.600
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 π.