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In this video, we will learn how to find indefinite integrals of functions that result in reciprocal trigonometric functions.
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We recall first that the derivative with respect to π₯ of the sec of π₯, or sec π₯, is equal to the sec of π₯ times the tan of π₯, or sec π₯ tan π₯.
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Recalling that integration is the inverse of differentiation, integrating both sides of this equation with respect to π₯ undoes the differentiation operation while also adding an arbitrary constant of integration.
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We can quote this as a standard result.
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The indefinite integral of sec π₯ tan π₯ with respect to π₯ is equal to sec π₯ plus an arbitrary constant πΆ.
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In our first example, we will use this formula to solve an indefinite integral.
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Determine the indefinite integral of nine tan three π₯ sec three π₯ with respect to π₯.
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The integrand here contains the product of a tangent and a secant, both of which have the same argument three π₯.
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We can recall the following standard integral.
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The indefinite integral of sec π₯ tan π₯ with respect to π₯ is equal to sec of π₯ plus πΆ.
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To use this result in this example, we need to modify the argument three π₯ of the trigonometric functions.
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We can do this by making a substitution.
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Weβll let π’ equal three π₯.
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It follows then that dπ’ by dπ₯ is equal to three.
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And whilst dπ’ by dπ₯ is not a fraction, we can treat it a little like one.
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So equivalently dπ’ is equal to three dπ₯, or one-third dπ’ is equal to dπ₯.
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We can now perform this substitution replacing three π₯ with π’ and dπ₯ with one-third dπ’ to obtain the indefinite integral of nine tan π’ sec π’ one-third dπ’.
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The constant in the integrand simplifies to three, and we can then bring this factor of three out the front to give three multiplied by the indefinite integral of tan π’ sec π’ with respect to π’.
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By the standard result, this integrates to three sec π’ plus a constant of integration πΆ.
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All that remains is to undo our substitution, so we need to replace π’ with three π₯, which gives our final answer to the problem.
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Weβve found that the indefinite integral of nine tan three π₯ sec three π₯ with respect to π₯ is equal to three sec three π₯ plus a constant of integration πΆ.
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In our next example, we will solve an indefinite integral which requires us to simplify the integrand first.
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Determine the indefinite integral of negative eight sec seven π₯ multiplied by negative four cos squared seven π₯ plus six tan seven π₯ with respect to π₯.
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Letβs begin by distributing the parentheses in the integrand.
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The integrand becomes 32 sec seven π₯ cos squared seven π₯ minus 48 sec seven π₯ tan seven π₯.
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Now since sec of seven π₯ is equal to one over cos of seven π₯, we can replace the first term with 32 cos squared seven π₯ over cos of seven π₯.
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We can cancel a factor of cos seven π₯ from the numerator and denominator, so the first term simplifies to 32 cos seven π₯.
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And now we need to determine the indefinite integral of 32 cos seven π₯ minus 48 sec seven π₯ tan seven π₯ with respect to π₯.
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The first term in the integrand involves a cosine function, and the second term involves a product of a secant and tangent function each with the same argument of seven π₯.
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We will therefore need to recall the following standard integrals.
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The integral of cos π₯ with respect to π₯ is equal to sin π₯ plus πΆ.
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And the integral of sec π₯ tan π₯ with respect to π₯ is equal to sec π₯ plus πΆ.
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However, before we can apply these formulas to solve the given indefinite integral, we need to modify the argument of the trigonometric functions.
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To do this, we can perform a substitution.
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We let π’ equal seven π₯, which in turn implies that dπ’ by dπ₯ is equal to seven.
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Now dπ’ by dπ₯ is not a fraction, but we can treat it a little like one.
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So equivalently one-seventh dπ’ is equal to dπ₯.
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Performing the substitution, we obtain the indefinite integral of 32 cos π’ minus 48 sec π’ tan π’ one-seventh dπ’.
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We can simplify this by splitting the integrand and taking each constant factor out the front to give 32 over seven multiplied by the indefinite integral of cos π’ dπ’ minus 48 over seven multiplied by the indefinite integral of sec π’ tan π’ dπ’.
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Applying the standard results, we obtain 32 over seven sin π’ plus a constant of integration πΆ one minus 48 over seven sec π’ plus a constant of integration πΆ two.
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We can combine the two constants of integration into a single arbitrary constant πΆ.
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Finally, we need to reverse the substitution by replacing π’ with seven π₯.
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And so we obtain our final answer, which is 32 over seven sin seven π₯ minus 48 over seven sec seven π₯ plus πΆ.
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Letβs consider another example involving the product of sec π₯ and tan π₯.
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Determine the indefinite integral of seven sec π₯ multiplied by tan π₯ minus five sec π₯ with respect to π₯.
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Since we have a factored expression in the integrand, weβll begin by distributing the parentheses, which gives the indefinite integral of seven sec π₯ tan π₯ minus 35 sec squared π₯ with respect to π₯.
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The first term is the product of a secant and tangent function, and the second is the square of the secant function.
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To solve this problem, we will need to recall two standard integrals.
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The indefinite integral of sec π₯ tan π₯ with respect to π₯ is equal to sec π₯ plus πΆ.
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And the indefinite integral of sec squared π₯ with respect to π₯ is equal to tan π₯ plus πΆ.
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Separating the integrand and taking the constant factors out the front of each integral, we obtain seven multiplied by the indefinite integral of sec π₯ tan π₯ with respect to π₯ minus 35 multiplied by the indefinite integral of sec squared π₯ with respect to π₯.
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We can now apply the standard results to this problem.
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Integrating, we obtain seven sec π₯ plus a constant of integration πΆ one minus 35 tan π₯ plus a constant of integration πΆ two.
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Combining the two arbitrary constants into a single constant πΆ, we obtain our final answer.
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The indefinite integral of seven sec π₯ multiplied by tan π₯ minus five sec π₯ with respect to π₯ is equal to seven sec π₯ minus 35 tan π₯ plus πΆ.
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In the examples weβve seen so far, weβve used the formula for indefinite integrals that give results involving the secant and tangent functions.
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Letβs now consider another type of indefinite integral involving a different reciprocal trigonometric function.
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Recall that the derivative of the csc of π₯, or csc π₯, with respect to π₯ is equal to negative the csc of π₯ times the cot of π₯.
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Integrating both sides with respect to π₯ as before, we obtain another standard integral.
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The indefinite integral of csc π₯ cot π₯ with respect to π₯ is equal to negative csc π₯ plus πΆ.
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Notice the similarity between this standard integral and the previous one.
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By replacing each trigonometric function in the integral by its complement, we obtain the same result, with the exception that we switch the negative sign to a positive one.
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This resemblance is not a coincidence.
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Recall that the complementary trigonometric functions cosine, cotangent, and cosecant take the complementary angles of their counterparts.
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In other words, cos of π₯ is equal to sin of π by two minus π₯, cot of π₯ is equal to tan of π by two minus π₯, and csc of π₯ is equal to sec of π by two minus π₯.
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Returning to our standard integral then, we can rewrite the indefinite integral of csc π₯ cot π₯ with respect to π₯ as the indefinite integral of sec of π by two minus π₯ tan of π by two minus π₯ with respect to π₯.
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We can then use the substitution π’ equals π by two minus π₯ such that dπ’ by dπ₯ is equal to negative one or equivalently negative dπ’ is equal to dπ₯.
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Making this substitution in the integral above, we obtain that the integral of csc π₯ cot π₯ with respect to π₯ is equal to the negative of the integral of sec π’ tan π’ with respect to π’.
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Using the standard integral of sec π’ tan π’ with respect to π’, we obtain negative sec π’ plus πΆ.
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And replacing π’ with π by two minus π₯, we obtain negative sec of π by two minus π₯ plus πΆ.
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And finally, switching both the complement argument and the complement function, we get negative csc π₯ plus πΆ.
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The inclusion of the negative sign compared to the two previous standard integrals results from the substitution of π’ equals π by two minus π₯, leading to negative dπ’ equals dπ₯, which then carries through the rest of the integration.
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Letβs now consider an example of an indefinite integral using this formula.
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Determine the indefinite integral of two csc three π₯ cot three π₯ with respect to π₯.
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The given integrand is the product of a cosecant and cotangent function, both with the argument of three π₯.
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We recall the standard integral of the product of the cosecant and cotangent functions.
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The indefinite integral of csc of π₯ multiplied by cot of π₯ with respect to π₯ is equal to the negative csc of π₯ plus πΆ.
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In the given integrand, the argument for both functions is three π₯ rather than π₯.
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So in order to apply the standard integral, we need to use a substitution.
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We let π’ equal three π₯, which in turn implies that dπ’ by dπ₯ is equal to three.
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And so one-third dπ’ is equivalent to dπ₯.
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Making this change of variable in the integral, we get the indefinite integral of two csc π’ cot π’ one-third dπ’.
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Taking the constant factor of two-thirds out the front of the integral and then applying our standard result gives negative two-thirds csc π’ plus a constant of integration πΆ.
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All that remains is to reverse the substitution by replacing π’ with three π₯.
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And so we obtain our final answer.
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The indefinite interval of two csc three π₯ cot three π₯ with respect to π₯ is equal to negative two-thirds csc of three π₯ plus πΆ.
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For our final formula, we want to determine the indefinite integral of csc squared of π₯ with respect to π₯.
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To start, we recall that the integral of sec squared π₯ with respect to π₯ is equal to tan π₯ plus πΆ.
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As discussed earlier, we can obtain the complementary counterpart to this result by replacing sec squared π₯ and tan π₯ with their complement, csc squared of π₯ and cot π₯.
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But then we must also include a factor of negative one on the right-hand side.
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This gives our final standard integral.
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The indefinite integral of csc squared of π₯ with respect to π₯ is equal to negative cot of π₯ plus πΆ.
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If ever we need to integrate tan squared π₯ or cot squared π₯, we can use the trigonometric identities tan squared of π₯ is equal to sec squared of π₯ minus one and cot squared of π₯ is equal to csc squared π₯ minus one to rewrite both of these integrals in terms of the standard integrals we already know.
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If you canβt remember either of these identities, they can each be derived from the Pythagorean identity of sin squared π₯ plus cos squared π₯ is identically equal to one.
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In our final example, we will evaluate an indefinite integral involving cot squared π₯ by first applying this trigonometric identity and second by applying the formula for the antiderivative of csc squared π₯.
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Determine the indefinite integral of negative five multiplied by the cot squared of four π₯ plus seven plus one with respect to π₯.
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To begin, since negative five is a constant, we can take it outside the integrand.
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Next, since the argument of the trigonometric function is four π₯ plus seven instead of π₯, we can make a substitution.
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We can let π’ equal four π₯ plus seven.
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And it then follows that dπ’ by dπ₯ is equal to four or equivalently one-quarter dπ’ is equal to dπ₯.
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Making this change of variable in the integral, we obtain negative five multiplied by the indefinite integral of cot squared π’ plus one one-quarter dπ’, which we can write as negative five over four multiplied by the indefinite integral of cot squared π’ plus one with respect to π’.
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The integrand contains the square of the cotangent function.
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The antiderivative of cot squared π₯ is not readily available, but we do know that cot squared π₯ can be expressed in terms of csc squared π₯, which we do know the antiderivative of.
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We recall that cot squared of π₯ is equal to csc squared of π₯ minus one.
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And this can be obtained from the Pythagorean identity.
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Hence, the integrand becomes csc squared of π’ minus one plus one or simply csc squared of π’.
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We can now proceed by recalling the standard result that the indefinite integral of csc squared of π₯ with respect to π₯ is equal to negative cot of π₯ plus πΆ.
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Applying this result, we obtain negative five over four multiplied by negative cot π’ plus πΆ.
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Reversing our substitution, so replacing π’ with four π₯ plus seven, we obtain our final answer, which is that the indefinite integral of negative five multiplied by cot squared of four π₯ plus seven plus one with respect to π₯ is equal to five over four multiplied by cot of four π₯ plus seven plus πΆ.
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Letβs now recap some of the key points from this video.
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Firstly, the indefinite integral of the product of the sec of π₯ and the tan of π₯ with respect to π₯ is equal to the sec of π₯ plus πΆ.
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To find the integral of a complementary trigonometric function, we replace every function with its complement function and change the sign of the antiderivative.
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For example, for the integral of the complementary function of the function in the first key point, the indefinite integral of the csc of π₯ multiplied by the cot of π₯ with respect to π₯ is equal to the negative csc of π₯ plus πΆ.
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And likewise, for the integral of the csc squared of π₯ with respect to π₯, we use our knowledge of the standard integral of the sec squared of π₯ with respect to π₯ to give the negative of the cot of π₯ plus πΆ.
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And finally, standard results for the integrals of the tan squared of π₯ and the cot squared of π₯ are not readily available, but we can use the trigonometric identities tan squared π₯ plus one is identically equal to sec squared π₯ and cot squared π₯ plus one is identically equal to csc squared π₯ to express any integrals of tan squared π₯ or cot squared π₯ in terms of the standard integrals of sec squared π₯ and csc squared π₯ that we already know.