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Use a trigonometric substitution to evaluate the integral of one over the square root of nine plus π₯ squared with respect to π₯.
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Weβre told to use a trigonometric substitution to evaluate this integral.
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But how do we decide which one we need?
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Well, there are three main situations we need to think about.
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If weβre looking to evaluate an integral that involves the square root of some number squared minus π₯ squared, we use the substitution π₯ equals that number multiplied by sin of π’.
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If weβre looking to integrate the square root of some number squared plus π₯ squared, we let our substitution be π₯ equals that number multiplied by tan π’.
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And finally, if weβre looking to integrate the square root of π₯ squared minus some number squared, we set our substitution as π₯ is equal to that number times sec π’.
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Our integral involves the square root of some number squared plus π₯ squared.
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So, weβre going to use the second situation.
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Nine is three squared, so the substitution weβre going to use is let π₯ equal three tan π’.
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Our next step is to differentiate this π₯ with respect to π’.
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The derivative of tan π’ is sec squared π’.
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So, differentiating three tan π’, and we get three sec squared π’.
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And this means dπ₯ must be equal to three sec squared π’ dπ’.
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Before we replace dπ₯ in our integral with three sec squared π’ dπ’, we need to decide how weβre going to express the square root of nine plus π₯ squared in terms of π’.
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Well, replacing π₯ with three tan π’, and this becomes the square root of nine plus three tan π’ squared.
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Three tan π’ all squared is the same as nine tan squared π’, so we can rewrite the square root of nine plus π₯ squared as shown.
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Now we can actually further simplify this by factoring nine in nine plus nine tan squared π’.
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And then, we can separate our roots.
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And it becomes the square root of nine times the square root of one plus tan squared π’.
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The square root of nine is three.
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And then, we use the identity one plus tan squared π’ equals sec squared π’.
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And this means the square root of one plus tan squared π’ is just sec π’.
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So, we can rewrite the square root of nine plus π₯ squared in our integral as three sec π’.
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And now weβre ready to make the substitutions.
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We need to integrate three sec squared π’ divided by three sec π’ with respect to π’.
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Notice how we can cancel these threes.
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We then see that sec squared π’ divided by sec π’ is simply sec π’.
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And our final step is to integrate sec π’ with respect to π’.
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And thereβs a general result for this.
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The integral of sec π’ with respect to π’ is ln of sec π’ plus tan π’ plus that constant of integration π.
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However, we were looking to evaluate our integral in terms of π₯.
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So, weβre going to need to find a way to substitute π₯ back into our expression.
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We go back to the original substitution.
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And we said let π₯ be equal to three tan π’.
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If we divide through both sides of this equation by three, we see that tan π’ is equal to π₯ over three.
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Similarly, we saw that we could replace the square root of nine plus π₯ squared with three sec π’.
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If we divide through by three here, we see that sec π’ is equal to the square root of nine plus π₯ squared all over three.
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And we can now replace tan π’ and sec π’ with expressions in terms of π₯.
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And weβve evaluated our integral.
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Itβs ln of the square root of nine plus π₯ squared over three plus π₯ over three plus that constant of integration π.