



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!
Dejavu! 





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. 





X^{2}
+ 2X + 5 


X^{2}
+ 2X + 1  1 + 5 


(X^{2}
+ 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 ... 





2sec^{2}qdq
= 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 ... 





sec^{2}qdq
= 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 
