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Determine the indefinite integral of nine tan three 𝑥 sec three 𝑥 with respect to 𝑥.
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The integrand here contains the product of a tangent and a secant, both of which have the same argument three 𝑥.
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We can recall the following standard integral.
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The indefinite integral of sec 𝑥 tan 𝑥 with respect to 𝑥 is equal to sec of 𝑥 plus 𝐶.
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To use this result in this example, we need to modify the argument three 𝑥 of the trigonometric functions.
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We can do this by making a substitution.
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We’ll let 𝑢 equal three 𝑥.
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It follows then that d𝑢 by d𝑥 is equal to three.
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And whilst d𝑢 by d𝑥 is not a fraction, we can treat it a little like one.
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So equivalently d𝑢 is equal to three d𝑥, or one-third d𝑢 is equal to d𝑥.
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We can now perform this substitution replacing three 𝑥 with 𝑢 and d𝑥 with one-third d𝑢 to obtain the indefinite integral of nine tan 𝑢 sec 𝑢 one-third d𝑢.
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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 𝑢 sec 𝑢 with respect to 𝑢.
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By the standard result, this integrates to three sec 𝑢 plus a constant of integration 𝐶.
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All that remains is to undo our substitution, so we need to replace 𝑢 with three 𝑥, which gives our final answer to the problem.
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We’ve found that the indefinite integral of nine tan three 𝑥 sec three 𝑥 with respect to 𝑥 is equal to three sec three 𝑥 plus a constant of integration 𝐶.