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Which of the following is the integration of the sec of π₯ multiplied by the tan of π₯?
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Is it option (A) the sec of π₯ plus π?
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Option (B) negative the csc of π₯ plus π.
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Option (C) negative the sec of π₯ plus π.
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Is it option (D) the csc of π₯ plus π?
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Or is it option (E) the cot of π₯ plus π?
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In this question, weβre 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₯.
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And the first thing we need to realize is our functions are given in terms of the variable π₯.
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So weβre going to need to integrate this with respect to π₯.
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We can actually answer this question in many different ways.
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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₯.
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And we can differentiate each of the five given options by recalling three of our derivative results involving trigonometric functions.
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We know the derivative of the sec of π₯ with respect to π₯ is the sec of π₯ times the tan of π₯.
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The derivative of the csc of π₯ with respect to π₯ is negative the csc of π₯ times the cot of π₯.
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And the derivative of the cot of π₯ with respect to π₯ is negative csc squared of π₯.
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This allows us to differentiate all five of the given options.
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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₯.
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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 πΆ.
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And this is a perfectly valid method for answering this question.
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However, it relies on one of two things.
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Either we need to be given the options to differentiate or we need to recall the corresponding derivative rule.
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These are valid methods of answering the question.
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However, it can be useful to do this directly.
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So letβs 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.
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To do this, weβre 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₯.
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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₯.
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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₯.
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And this is not an easy integral to evaluate.
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So weβre going to simplify this integral by using a substitution.
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Weβre going to use the substitution π’ is equal to the cos of π₯.
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And we recall whenever we integrate by using substitution, we need to find an expression for the differentials.
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To do this, weβre going to need to differentiate our substitution π’ with respect to π₯.
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We know that the derivative of the cos of π₯ with respect to π₯ is negative the sin of π₯.
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This gives us that dπ’ by dπ₯ is equal to negative the sin of π₯.
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And now we know that dπ’ by dπ₯ is not a fraction.
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However, we can treat it a little bit like a fraction when weβre using integration by substitution.
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This will allow us to find an equation for the differentials.
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This gives us that dπ’ is equal to negative sin of π₯ dπ₯.
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We now want to use our π’-substitution to rewrite the integral.
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First, in the denominator of our integral, weβll substitute π’ for the cos of π₯.
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This gives us a new denominator of our integrand of π’ squared.
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Second, we want to use our expression for the differentials.
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However, we can see that negative sin of π₯ dπ₯ does not appear in the integral.
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Instead, we have sin of π₯ dπ₯.
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So weβre going to need to rewrite our integral by multiplying our integrand by negative one and the entire integral by negative one.
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This allows us to replace negative sin of π₯ dπ₯ with dπ’.
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This gives us negative the integral of one over π’ squared with respect to π’.
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And we can now evaluate this integral by using the power rule for integration.
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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 πΆ.
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We add one to our exponent of π’ and then divide by this new exponent.
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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.
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We need to add one to our exponent of π’ and divide by the new exponent.
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This gives us π’ to the power of negative two plus one over negative two plus one.
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And remember, we need to multiply this value by negative one and πΆ is a constant.
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So multiplying this value by negative one wonβt change the fact that itβs constant.
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So we can just add πΆ at the end of this expression.
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So we have negative one times π’ to the power of negative two plus one over negative two plus one plus πΆ.
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And we can simplify this expression by noting negative two plus one is equal to negative one.
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This gives us a shared factor of negative one in the numerator and denominator.
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So weβre just left with π’ to the power of negative one plus πΆ.
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And of course we know that π’ to the power of negative one is the same as one over π’.
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So we have one over π’ plus πΆ.
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Finally, since our original integral was given in terms of π₯, we should write our answer in terms of π₯.
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We can do this by using our substitution.
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π’ is equal to the cos of π₯.
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This gives us one over the cos of π₯ plus πΆ.
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But remember, one over the cos of π₯ is equal to the sec of π₯.
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So we can rewrite this as the sec of π₯ plus πΆ, which is our final answer.
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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₯.
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And in all of these cases, we were able to show itβs equal to the sec of π₯ plus πΆ, which is given as option (A).