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Triangles Part II, The Revenge
Integration by Trig Substitution

 

 The stuff we have talked about up to now

 will let you solve most integrals.
 There are a few more tricks to show you
 that may help for really nasty problems.
 Like maybe this one ...
 

 
 What can we do with it?
 
 We can't do integration by parts on it.
 
 We don't have an inside function derivative sitting there
 to "stuff" back into another part.
 That means we can't do a "u and du" substitution.
 
 What do we do?
 
 We borrow an idea from trigonometry.
 
 We need to turn our big nasty, messy term
 into one simple term.
 The trick we use comes from the parts of a right triangle.
 For this particular problem, 
 we use this particular right triangle ...
 

 
 The first question is, what is the length of the third side?
 
 To find it, we need some help from our old buddy Pythagoras ...
 

 
 Wow! Deja-vu!
 
 But so what?
 What does that get us?
 
 Here's the deal.
 Let's take our triangle and work out the six trig ratios.
 

 
 Now let's look at the sec q one ...
 

 
 multiply both sides by 1/3 ...
 

 
 We couldn't find an integral for that nasty root term,
 but maybe we'll have better luck with the secant.
 Hmmm, something tells me we will.
 Otherwise, this page wouldn't exist!
 
 We substitute in 1/3 secq for the nasty term:
 

 
 Next we need to deal with the dX part.
 We need to turn it into something with q too.
 
 Let's look at the sine from our triangle ...
 

 
 Multiply both sides by 3 ...
 

 3sinq = X

 
 That's close, but we need the derivative of X (dX),
 so we take the derivative of both sides ...
 

 3cosq = dX

 
 And now we can substitute for dX ...
 

 
 Simplify,
 remember secq = 1/cosq ...
 

 
 That's great, but we have one big problem.
 We started with X's and now we have q's.
 
 To solve the problem, we go back to the tig ratios.
 In particular, the sine one ...
 

 
 Watch carefully ...
 

 
 What was that???
 
 It's called the inverse sine (sometimes called the arcsine).
 It means "the angle that has X/3 for a sine is q"
 
 An example would be ...
 

 

 
 If this is still fuzzy,
 you should probably review trig a bit.
 
 Go ahead, we'll wait right here for you.
 
 Back already? That was quick!
 OK, back to the problem.
 

 
 So we substitute this into the equation ...
 

 
 I know it looks a bit goofy, but that's the answer.
 
 There are lots of problems where this trig substitution stuff works.
 
 In general, if you have an integral of X 2 + a, X 2 - a, or a - X 2
 (where a is any number)
 and you don't have what you need to do a u and du substitution, 
 you can use this trick.
 
 The triangles and trig functions for these three problem types are ...
 

 

 WARNING!    WARNING!    DANGER!    DANGER!

 
 Don't try to memorize all this junk!
 Learn how to build the part you need when you need it.
 You know Pythagoras (a 2 + b 2 = c 2).
 Remember that the three sides of the triangle
 are an X, a number, and a root.
 Then look at what you are dealing with in the problem
 and decide which of the three you need.
 
 Let's try another one ...
 
 Example:
 

 
 You might not recognize this one right off
 as a trig substitution problem.
 It doesn't have a root radical,
 and it has a "middle" term (the 2X).
 It is one, believe me.
 
 To see it, we first do something called "complete the square"
 on the denominator.
 
 X2 + 2X + 5
 X2 + 2X + 1 - 1 + 5
(X2 + 2X + 1) - 1 + 5
(X + 1)2 + 4
 

 That gives us ...

 
 
 Getting closer?
 Substitute: u = X + 1, and get ...
 

 
 Close enough? No? OK, how about ...
 

 
 Now the question is, which of the three triangles do we need?
 

 
 The term in the problem is like
 That means we need triangle (A).
 So we have ...
 

 
 Our problem is ...
 

 
 So we need the cosine: a 2 = 4, so a = 2
 

 
 Substituting this into our integral, 
 we get ...
 

 
 To get dq, we use tanq ...
 

 
 Take the derivative of both sides ...
 

 2sec2qdq = dX

 
 Substitute this for dX ...
 

 
 And ...
 

 
 So ...
 

 
 Now let's get back to X's ...
 

 
 So we have ...
 

 
 These problems don't always show up as fractions ...
 
 Example:
 

 
 The triangle we need is ...
 

 
 So we use ...
 

 

 
 and we use ...
 

 tanq = X

 
 to find dX, take the derivative of both sides ...
 

 sec2qdq = dX

 
 Substituting these, we get ...
 

 

 
 Now we need the secant reduction formula ...
 

 
 So we have ...
 

 
 Now we get back to X's using ...
 

 
 So we get ...
 

 
 Which means ...
 

 
 Who would have thought this little integral,
 would make such a mess!
 

   copyright 2005 Bruce Kirkpatrick

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