4&*('&%/$z#" 10.B lkBhcTrig substitutionskBhcThere are number of special forms that suggest a trig substitution. The most common candidates for trig substitutions include the formstrqSWkBhcaopxEkBhc2kBhc  - xopxEkBhc2kBhc which suggests kBhc x = a sin trqSWkBhcaopxEkBhc2kBhc  + xopxEkBhc2kBhc which suggests kBhc x = a tan trqSWkBhcxopxEkBhc2kBhc  - aopxEkBhc2kBhc which suggests kBhc x = a sec kBhc=6Here are some examples where these substitutions help.\kBhc&In the first example we computeoetnIqtrqS4kBhc 1 - xopxEkBhc2kBhcxkBhc1kBhc -1kBhcThis example has an interesting interpretation. What we are computing here is the area of a semicircle of radius 1. We know in advance that the answer should be kBhc /2kBhc.kBhc4-The recommended substitution in this case is .kBhc x = sin #kBhc dx = cos d"kBhc' Applying this substitution givesetnIXtrqS4kBhc 1 - xopxEkBhc2kBhcxkBhc  = etnIxtrqSCkBhc1 - sinopxEkBhc2kBhc  kBhc  cos kBhckBhc  = etnISkBhc cosopxEkBhc2kBhc  kBhcrkBhcD=We can solve the cosine squared integral via the substitution kBhc cosopxEkBhc2kBhc  = carFDkBhc1 + cosGrapkBhc 2 kBhc2kBhc.q etnISkBhc cosopxEkBhc2kBhc  kBhckBhc  = etnIhcarFDkBhc1 + cosGrapkBhc 2 kBhc2kBhckBhc  = carF$kBhc1kBhc2kBhc  + carF$kBhc1kBhc4kBhc  sinGrapkBhc 2 kBhc  + CkBhc/(The last step is to substitute back for kBhckBhc by using kBhc = sinopxEkBhc -1kBhc  xkBhc:carF$kBhc1kBhc2kBhc  + carF$kBhc1kBhc4kBhc  sinGrapkBhc 2 kBhc  = carF$kBhc1kBhc2kBhc  sinopxEkBhc -1kBhc  x + carF$kBhc1kBhc4kBhc  sinGrap>kBhc 2 sinopxEkBhc -1kBhc  xkBhc  + CkBhc70Substituting the endpoints and simplifying givesp GrapcarF$kBhc1kBhc2kBhc  sinopxEkBhc -1kBhc  1 + carF$kBhc1kBhc4kBhc  sinGrap>kBhc 2 sinopxEkBhc -1kBhc  1kBhc  - Grap carF$kBhc1kBhc2kBhc  sinopxEkBhc -1kBhc GrapkBhc -1kBhc  + carF$kBhc1kBhc4kBhc  sinGrapVkBhc 2 sinopxEkBhc -1kBhc GrapkBhc -1kBhc  = carF$kBhckBhc4kBhc  - Grap=kBhc - carF$kBhckBhc4kBhc  = carF$kBhckBhc2skBhc#This is the expected result.kBhc A more difficult integralnkBhcThe next example isetnIXtrqS4kBhc 1 + xopxEkBhc2kBhcxZkBhc$The suggested substitution isckBhc x = tan kBhcwhich leads tofkBhc d x = secopxEkBhc2kBhc  d kBhc6/Substituting these back into the integral givesdetnItrqSCkBhc1 + tanopxEkBhc2kBhc  kBhc  secopxEkBhc2kBhc  kBhcbkBhc = etnItrqS?kBhc secopxEkBhc2kBhc  ukBhc  secopxEkBhc2kBhc  kBhcgkBhc = etnISkBhc secopxEkBhc3kBhc  kBhcekBhcyYou can solve this integral through a clever application of integration by parts. The trick is to rewrite the integral asetnISkBhc secopxEkBhc3kBhc  kBhcukBhc  = etnIYkBhc secopxEkBhc2kBhc sec kBhc_kBhcand integrate the kBhc secopxEkBhc2kBhc  kBhc' term while differentiating the kBhc sec kBhc  term.etnITGrapkBhc sec opxEkBhc3kBhckBhc = sec tan - etnIYkBhc sec tanopxEkBhc2kBhc  kBhc`kBhc>7We can evaluate the latter integral by a trig identity.etnIYkBhc sec tanopxEkBhc2kBhc  kBhckBhc  = etnItkBhc sec Grap?kBhc secopxEkBhc2kBhc  - 1kBhckBhc = etnIhkBhc -sec + GrapkBhc sec opxEkBhc3kBhckBhc ThusetnITGrapkBhc sec opxEkBhc3kBhckBhc = sec tan + etnI0kBhc sec kBhckBhc  - etnITGrapkBhc sec opxEkBhc3kBhckBhc!Rearranging slightly gives3kBhc 2 etnITGrapkBhc sec opxEkBhc3kBhckBhc = sec tan + etnI0kBhc sec kBhc9kBhcEarlier we determined-etnI0kBhc sec kBhckBhc  = ln VsbA kBhc sec + tan kBhc  + C>kBhc Hence?etnITGrapkBhc sec opxEkBhc3kBhckBhc  = carF$kBhc1kBhc2kBhc GrapWkBhcsec tan + ln VsbA kBhc sec + tan kBhc  + C<kBhc2+The last step is to substitute back in for kBhcxkBhc::kBhc x = tan )kBhc = tanopxEkBhc -1kBhc  x etnIXtrqS4kBhc 1 + xopxEkBhc2kBhcxkBhc  = carF$kBhc1kBhc2kBhc GrappkBhc secGrap<kBhc tanopxEkBhc -1kBhc  xkBhc  tanGrap<kBhc tanopxEkBhc -1kBhc  xkBhc  + ln VsbAkBhc secGrap<kBhc tanopxEkBhc -1kBhc  xkBhc  + tanGrap<kBhc tanopxEkBhc -1kBhc  xkBhc  + CkBhcIBOur last problem is figuring out how to simplify expressions like kBhc secGrap<kBhc tanopxEkBhc -1kBhc  xkBhc*#. Instead of substituting back for kBhckBhcG@ we can try a different approach. Writing the answer in the formcarF$kBhc1kBhc2kBhc GrapWkBhcsec tan + ln VsbA kBhc sec + tan kBhc  + CkBhc&we see that we have to compute kBhc sec kBhc given that kBhc tan = xkBhc^W. The key to handling situations like this is to go back to the original substitution (kBhc x = tan ) kBhcIBand interpret it as a statement about a right triangle with angle kBhckBhcJC. We can construct a right triangle with an angle whose tangent is kBhcxkBhc:3 by making the side opposite the angle have length kBhcxkBhcd] and the side adjacent to the angle have length 1. This forces the hypotenuse to have length trqS4kBhc 1 + xopxEkBhc2kBhc..warDRGzQ@{Gz?M xyairTloףp= @6iͿz(\@z6iͿz(\@0z@@Qx@ tQѿkBhc1(\@/b?Gz?`GzĿkBhcx a.V`ѿ*^@trqS4kBhc 1 + xopxEkBhc2Croh Ctrv <Q?kBhc-kBhc2+We can then read off from this diagram thatkBhc tan = x-kBhcsec = trqS4kBhc 1 + xopxEkBhc2kBhc Thus kBhc etnIXtrqS4kBhc 1 + xopxEkBhc2kBhcxkBhc  = carF$kBhc1kBhc2kBhc GrapWkBhcsec tan + ln VsbA kBhc sec + tan kBhc  + CkBhc = carF$kBhc1kBhc2GrapkBhc x trqS4kBhc 1 + xopxEkBhc2kBhc  + lnVsbAOtrqS4kBhc 1 + xopxEkBhc2kBhc  + xkBhc  + C-kBhcAnother triangle example-kBhcThe substitutions suggested above really come in useful in integrals in which those square root form appears in combination with other algebraic expressions. Consider this example.,etnIcarFhtrqS4kBhc 4 - xopxEkBhc2kBhcxopxEkBhc2kBhcx+kBhc1*The suggested substitution in this case is*kBhc x = 2 sin -kBhcd x = 2 cos d)kBhc81Making this substitution converts the integral to(etnIcarFtrqSEkBhc 4 - 4 sinopxEkBhc2kBhc  kBhc 4 sinopxEkBhc2kBhc  kBhc 2 cos kBhckBhc  = etnIcarFkBhc 2 trqSCkBhc1 - sinopxEkBhc2kBhc  kBhc 4 sinopxEkBhc2kBhc  kBhc 2 cos kBhckBhc  = etnIcarFrkBhc cosopxEkBhc2kBhc  kBhc sinopxEkBhc2kBhc  kBhckBhc 'kBhc = etnIRkBhc tanopxEkBhc2kBhckBhckBhc  = etnIWkBhc secopxEkBhc2kBhc  - 1kBhckBhc = tan - + C&kBhcOnce again the problem at the end is to reverse the substitution. To do this, we can replay the argument we used in the last example with a triangle constructed to ensure that kBhc sin = x/2kBhc.%warD} GzQ@{Gz?M xyairTlףp= p!6iͿ!(\@P6iͿP!(\@p!@@Qx@ tQѿtrqS4kBhc 4 - xopxEkBhc2(\@/b?Gz?`GzĿkBhcx a.V`ѿ*^@kBhc2Croh Ctrv <Q?kBhc-kBhc*#We read off from this diagram that kBhc tan = x/trqS4kBhc 4 - xopxEkBhc2kBhc:$etnIcarFhtrqS4kBhc 4 - xopxEkBhc2kBhcxopxEkBhc2kBhcxkBhc  = carFPkBhcxtrqS4kBhc 4 - xopxEkBhc2kBhc  - sinopxEkBhc -1Grap0carF$kBhcxkBhc2kBhc  + C!kBhc-&Extra algebra work is sometimes needed kBhcThe next example shows that sometimes we will have to do some preliminary algebra and a preliminary substitution before we can apply the trig substitution of our choice. Here is the problem:etnIcarFekBhc 2 xtrqSGkBhc 2 xopxEkBhc2kBhc + 3 x + 2kBhcxkBhcngThe form of the expression in the radical suggests that we should use the substitution appropriate for kBhcxopxEkBhc2kBhc  + aopxEkBhc2kBhc , which is kBhc x = a tan kBhcx. However, before we can apply that substitution, we have to make the expression in the radical look more like the form kBhcxopxEkBhc2kBhc  + aopxEkBhc2kBhcOH. The first thing to do is to eliminate the factor of 2 in front of the kBhcxopxEkBhc2kBhcXQ term. We can do this by factoring out a factor of 2 from underneath the radical.etnIcarFekBhc 2 xtrqSGkBhc 2 xopxEkBhc2kBhc + 3 x + 2kBhcxkBhc  = carF4kBhc1trqSkBhc2kBhc etnIcarFekBhc 2 xtrqSGkBhcxopxEkBhc2kBhc + 3/2 x + 1kBhcxkBhc6/The next step is to get rid of the superfluous kBhc 3/2 xkBhc{t term in the radical expression. The appropriate way to accomplish that is to complete the square in the polynomial.kBhcxopxEkBhc2kBhc + 3/2 x + 1 = xopxEkBhc2kBhc  + 2 GrapkBhc 3/4kBhc x + 1 = xopxEkBhc2kBhc  + 2 GrapkBhc 3/4kBhc  x + GrapkBhc 3/4opxEkBhc2kBhc  - GrapkBhc 3/4opxEkBhc2kBhc  + 1kBhc = GrapkBhcx + 3/4opxEkBhc2kBhc + 7/16kBhcC