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Problem 37 Hard Difficulty

Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent.

$ \displaystyle \int \sec^4 x\ dx $


$\frac{1}{3} \tan x \sec ^{2} x+\frac{2}{3} \tan x+C$


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Video Transcript

Okay. This question wants us to use technology to find this integral and compare it with the tabular answer. So if I plug this into Wolfram, Alfa gives us this answer for anti derivative. But if we use the table, we get this answer. So they clearly don't match. Although they do look kind of close. So let's see if we can manipulate the Wolfram MTA form into the tap their form. So let's start with tangent. Cubed X over negative six plus tangent Axe over too Plus 1/2 tangent X c can't squared of X So first, what we're gonna d'oh is remember one of our trig identities. So tangents squared. X plus one equals C can't squared x So that means that tangent squared of X equals C can't squared X minus one. So we're going to use this negative so we can factor this as Tanja necks times, tangents, squared acts over six and the rest of the terms were not going to mess with. So now we're gonna substitute our expression in for tangents squared so you get negative tangent squared or negative tender necks Still times c can't squared X minus one over six plus tangent X over too plus 1/2 c can't square Dex, Tanja and acts so that turns into negative 16 Tanja necks c can't squared X plus 1/6 tangent acts plus 1/2 tanja necks plus 1/2 see Can't squared Tanja necks So now we're pretty much done We just need to collect our like terms So we get negative 16 plus 1/2 c can't square Dax Tangin axe plus 1/6 plus 1/2 Tanja necks plus c So now all they have to do is simplify our coefficients and we get well, 36 You're sorry? Yeah, 3/6 minus 1/6 is 1/3. So we get C can't square Dax Tangin axe over three plus 36 plus 1/6 is 2/3 Tanja necks plus c. So this is our new form and let's compare this toe are tabular answer And that's the exact form that we got. So the integral expressions are equivalent after all,

University of Michigan - Ann Arbor
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