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7 .4, problem 37.
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So the integral of square to 9x squared minus 25 over x cubed dx.
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So in order to work with this, i need to do a little bit of algebraic manipulation first because i need that quadratic term to have a coefficient of 1.
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So let's just factor out the 9.
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So that can be written as 9x squared minus 25 over 9.
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And then the square root of 9 is 3.
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So this is x squared minus 25 over 9.
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Now based on this, i know the trig substitution is a secant substitution.
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So let x equal 5 thirds secant of theta.
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Then dx is equal to 5 thirds, secant theta, tangent theta, d theta.
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And then this square root term, so the square root of x squared minus 25 over 9, that's going to be 25 over 9, secant squared minus 1, so that's going to end up being 5 thirds tangent theta.
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Now let's substitute all of this in to transform this integral now in terms of theta.
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So this is going to be 5 thirds tangent of theta divided by xq, so that's going to be 5 thirds cubed and secant cubed of theta.
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And then dx is going to be 5 thirds, secant theta, tangent theta, d theta.
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So i see that i'm going to end up now with 1 over 5 thirds, so that's going to be 3 fifths.
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And i need to back up just a moment.
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When i got to this particular spot here there was a three that should be there so there should be a three in front of this integral so so if i do all of this this i end up with three times five three -fifths is going to be nine -fifths so this answer is going to be nine -fifths the integral and you're going to have the tangent squared of theta divided by the secant squared of theta d theta.
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So let's write this in terms of sine and cosine.
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So this is nine -fifths.
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This is sine squared theta over cosine squared theta.
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And secant squared is just one over the cosine d theta...