Question Video: Integrating a Reciprocal Trigonometric Function Mathematics

Determine β«(1/(1 β cosΒ² 2π₯)) dπ₯.

03:33

Video Transcript

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

In this question, weβre asked to evaluate the integral of a trigonometric function. And we can see this is a difficult function to integrate. Itβs not a standard function we know how to integrate, nor is it in a form which we can integrate already. So weβre going to need to rewrite our integral in a way we can integrate. And thereβs a few different ways of doing this. Since this is a trigonometric function, weβre going to start by trying to rewrite our integrand by using trigonometric identities. And we can notice something interesting about our integrand. The denominator involves one minus the cosine squared function. And this can remind us of the Pythagorean identity.

We recall this tells us for any angle π, the sin squared of π plus the cos squared of π is equal to one. In particular, if we subtract the cos squared of π from both sides of the equation and rewrite π equal to two π₯, we get that one minus the cos squared of two π₯ is equal to the sin squared of two π₯. We can use this to rewrite our integrand, giving us the integral of one over one minus the cos squared of two π₯ with respect to π₯ is equal to the integral of one over the sin squared of two π₯ with respect to π₯.

We can then simplify our integrand even further by using one of our reciprocal trigonometric identities. We recall the csc of π is equal to one divided by the sin of π. And our integrand is one divided by the sin squared of two π₯. So we can square both sides of the identity and replace π with two π₯. This then gives us the integral of the csc squared of two π₯ with respect to π₯. And now, this is almost in one of our standard integral results. We can recall the integral of the csc squared of π with respect to π is equal to negative the cotan of π plus the constant of integration πΆ.

However, weβre not taking the csc squared of π₯; weβre taking the csc squared of two π₯. So weβre going to need to rewrite our integral, and weβll do this by using the π’-substitution, π’ is equal to two π₯. To integrate by substitution, we need to find an expression for the differentials. And we do this by differentiating our π’-substitution with respect to π₯. We get dπ’ by dπ₯ is the derivative of two π₯ with respect to π₯. This is the coefficient of π₯, which is two. Now, although dπ’ by dπ₯ is not a fraction, we can treat it a little bit like a fraction when weβre using integration by substitution. This will allow us to find an equation involving the differentials. We get one-half dπ’ is equal to dπ₯.

We can now use our π’-substitution to rewrite our integral. We have π’ is two π₯ and dπ₯ is a half dπ’. This gives us the integral of the csc squared of π’ times a half with respect to π’. And weβre almost ready to evaluate our integral. Weβll start by taking the constant factor of a half outside of the integral. This gives us a half times the integral of the csc squared of π’ with respect to π’. Now, we could just evaluate this integral by using our integral result. The integral of the csc of π’ with respect to π’ is negative the cotan of π’ plus the constant of integration. So we get one-half times negative the cotan of π’. And we need to add a half multiplied by a constant, but this is still a constant. So we can just add a constant of integration πΆ at the end of this expression.

We can simplify this expression by distributing the one-half over our parentheses. This gives us negative a half the cotan of π’ plus πΆ. And since our original integral was in terms of the variable π₯, we should give our answer in terms of π₯. We can do this by using our π’-substitution. This then gives us our final answer. The integral of one over one minus the cos squared of two π₯ with respect to π₯ is negative a half the cotan of two π₯ plus the constant of integration πΆ.

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