Question Video: Integrating a Product of Trigonometric Functions Mathematics

Which of the following is the integration of sec π₯ tan π₯? [A] sec π₯ + π [B] βcsc π₯ + π [C] βsec π₯ + π [D] csc π₯ + π [E] cot π₯ + π

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Video Transcript

Which of the following is the integration of the sec of π₯ multiplied by the tan of π₯? Is it option (A) the sec of π₯ plus π? Option (B) negative the csc of π₯ plus π. Option (C) negative the sec of π₯ plus π. Is it option (D) the csc of π₯ plus π? Or is it option (E) the cot of π₯ plus π?

In this question, weβre asked to determine which of five given options is the integral of the product of two given trigonometric functions, the sec of π₯ multiplied by the tan of π₯. And the first thing we need to realize is our functions are given in terms of the variable π₯. So weβre going to need to integrate this with respect to π₯. We can actually answer this question in many different ways. For example, since we know integration finds the most general antiderivative of a function, we can differentiate each of the five given options to determine which of these is an antiderivative of the sec of π₯ multiplied by the tan of π₯. And we can differentiate each of the five given options by recalling three of our derivative results involving trigonometric functions.

We know the derivative of the sec of π₯ with respect to π₯ is the sec of π₯ times the tan of π₯. The derivative of the csc of π₯ with respect to π₯ is negative the csc of π₯ times the cot of π₯. And the derivative of the cot of π₯ with respect to π₯ is negative csc squared of π₯. This allows us to differentiate all five of the given options. In particular, this allows us to see that the sec of π₯ is an antiderivative of the sec of π₯ times the tan of π₯. In terms of integrals, this means the integral of the sec of π₯ times the tan of π₯ with respect to π₯ is the sec of π₯ plus a constant of integration πΆ. And this is a perfectly valid method for answering this question.

However, it relies on one of two things. Either we need to be given the options to differentiate or we need to recall the corresponding derivative rule. These are valid methods of answering the question. However, it can be useful to do this directly. So letβs try and evaluate the integral of the sec of π₯ multiplied by the tan of π₯ with respect to π₯ by using our properties and results of integration. To do this, weβre going to start by rewriting our integrand by using the fact that the sec of π₯ is equal to one over the cos of π₯ and the tan of π₯ is equal to the sin of π₯ divided by the cos of π₯. This gives us the integral of one over the cos of π₯ multiplied by the sin of π₯ divided by the cos of π₯ with respect to π₯. We can then simplify our integrand to get the integral of the sin of π₯ divided by the cos squared of π₯ with respect to π₯.

And this is not an easy integral to evaluate. So weβre going to simplify this integral by using a substitution. Weβre going to use the substitution π’ is equal to the cos of π₯. And we recall whenever we integrate by using substitution, we need to find an expression for the differentials. To do this, weβre going to need to differentiate our substitution π’ with respect to π₯. We know that the derivative of the cos of π₯ with respect to π₯ is negative the sin of π₯. This gives us that dπ’ by dπ₯ is equal to negative the sin of π₯. And now we know that dπ’ by dπ₯ is not a fraction. However, 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 for the differentials. This gives us that dπ’ is equal to negative sin of π₯ dπ₯.

We now want to use our π’-substitution to rewrite the integral. First, in the denominator of our integral, weβll substitute π’ for the cos of π₯. This gives us a new denominator of our integrand of π’ squared. Second, we want to use our expression for the differentials. However, we can see that negative sin of π₯ dπ₯ does not appear in the integral. Instead, we have sin of π₯ dπ₯. So weβre going to need to rewrite our integral by multiplying our integrand by negative one and the entire integral by negative one. This allows us to replace negative sin of π₯ dπ₯ with dπ’. This gives us negative the integral of one over π’ squared with respect to π’. And we can now evaluate this integral by using the power rule for integration.

And we recall the power rule for integration tells us for any real constant π not equal to negative one, the integral of π’ to the power of π with respect to π’ is equal to π’ to the power of π plus one divided by π plus one plus a constant of integration πΆ. We add one to our exponent of π’ and then divide by this new exponent. To apply the power rule for integration, we need to use our laws of exponents to rewrite the integrand as π’ to the power of negative two. We need to add one to our exponent of π’ and divide by the new exponent. This gives us π’ to the power of negative two plus one over negative two plus one.

And remember, we need to multiply this value by negative one and πΆ is a constant. So multiplying this value by negative one wonβt change the fact that itβs constant. So we can just add πΆ at the end of this expression. So we have negative one times π’ to the power of negative two plus one over negative two plus one plus πΆ. And we can simplify this expression by noting negative two plus one is equal to negative one. This gives us a shared factor of negative one in the numerator and denominator. So weβre just left with π’ to the power of negative one plus πΆ. And of course we know that π’ to the power of negative one is the same as one over π’. So we have one over π’ plus πΆ.

Finally, since our original integral was given in terms of π₯, we should write our answer in terms of π₯. We can do this by using our substitution. π’ is equal to the cos of π₯. This gives us one over the cos of π₯ plus πΆ. But remember, one over the cos of π₯ is equal to the sec of π₯. So we can rewrite this as the sec of π₯ plus πΆ, which is our final answer.

Therefore, we were able to show several different ways of determining the integral of the sec of π₯ times the tan of π₯ with respect to π₯. And in all of these cases, we were able to show itβs equal to the sec of π₯ plus πΆ, which is given as option (A).