### Video Transcript

Determine the integral of the cos of two π₯ minus three divided by one minus cos squared of two π₯ minus three with respect to π₯.

In this question, weβre asked to evaluate the integral of a trigonometric expression. And we can see this is a very difficult expression to integrate. We canβt evaluate this integral directly. So weβre either going to need to apply an integral result, or weβre going to need to rewrite our integrand into a form which we can integrate. And since weβre working with trigonometric functions, itβs usually a good idea to try and rewrite our integrand first, because thereβs a lot of different ways of writing trigonometric expressions.

For example, we can notice the denominator of our integrand is almost in the form of the Pythagorean identity. The sin squared of π plus the cos squared of π is equal to one for any value of π. We can subtract the cos squared of π from both sides of the equation and then rewrite π to be two π₯ minus three. This gives us the sin squared of two π₯ minus three is equal to one minus the cos squared of two π₯ minus three. Therefore, we can use this to rewrite the denominator of our integrand. This then gives us the integral of the cos of two π₯ minus three divided by the sin squared of two π₯ minus three with respect to π₯.

And this is still not in a form which we can integrate easily. So instead, weβre going to need to keep rewriting our integrand. We can do this by noticing both the cosine function in our numerator and the sine function in our denominator have the same argument. This means we can rewrite our integrand by using any of our trigonometric identities. In particular, the cos of π divided by the sin of π is the cot of π. So if we replace π in this identity with two π₯ minus three, we can rewrite our integrand in terms of the cot of two π₯ minus three.

However, before we do this, remember, we have sin squared in our denominator, so we would still have an extra factor of one over the sin of two π₯ minus three. So we can also rewrite this in a similar manner. Remember, one divided by the sin of π is equal to the csc of π. Therefore, one over the sin of two π₯ minus three is equal to the csc of two π₯ minus three. Applying both of these identities to our integrand, we were able to rewrite our integral as the integral of the cot of two π₯ minus three multiplied by the csc of two π₯ minus three with respect to π₯.

And now, this integral is almost exactly in the form of one of our integral results. We can recall the integral of the cot of π times the csc of π with respect to π is negative the csc of π plus a constant of integration πΆ. However, our argument is not a single variable. Instead, itβs two π₯ minus three. So weβre going to need to rewrite our integral by using a substitution. To integrate by substitution, we need to find an expression for the differentials. So we need to differentiate π’ with respect to π₯. We can see dπ’ by dπ₯ is the coefficient of π₯, which in this case is two.

And although dπ’ by dπ₯ is not a fraction, we can treat it a little bit like a fraction when weβre using integration by substitution. This gives us the equation in terms of differentials one-half dπ’ is equal to dπ₯. Weβre now ready to apply our π’ substitution to help us evaluate this integral. We get the integral of the cot of π’ times the csc of π’ multiplied by one-half with respect to π’. And of course, we can take the factor of one-half outside of our integral.

Now, we can evaluate this integral by just applying our integral result. We get one-half multiplied by negative the csc of π’ plus a constant of integration πΆ. And we can simplify this expression. We can distribute one-half over the parentheses. And since our original integral was given in terms of π₯, we should give our answer in terms of π₯. Weβll do this by using our π’ substitution. This gives us negative one-half the csc of two π₯ minus three plus πΆ, which is our final answer.

Therefore, by using various trigonometric identities to rewrite our integral and then applying one of our integral results in π’ substitution, we were able to show the integral of the cos of two π₯ minus three divided by one minus the cos squared of two π₯ minus three with respect to π₯ is equal to negative a half the csc of two π₯ minus three plus πΆ.