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calculus 2 trigonometric substitutions

by: Alejandra D. Rocha

calculus 2 trigonometric substitutions 2314

Marketplace > University of Texas at El Paso > Math > 2314 > calculus 2 trigonometric substitutions
Alejandra D. Rocha
GPA 3.1

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About this Document

These notes cover integrals using trigonometric substitutions
Calculus 2
Class Notes
trigsubstitutions, Math, calculus2, integralswithtrigonometricfunctions, Sine, Cosine, Tangent, squarerootsintegrals
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This 15 page Class Notes was uploaded by Alejandra D. Rocha on Friday June 10, 2016. The Class Notes belongs to 2314 at University of Texas at El Paso taught by in Summer 2016. Since its upload, it has received 5 views. For similar materials see Calculus 2 in Math at University of Texas at El Paso.


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Date Created: 06/10/16
Trigonometric substitution Let’s start off with a couple of integrals that we should already be able to do with a standard  substitution.     Both of these used the substitution   and at this point should be pretty easy for  you to do.  However, let’s take a look at the following integral.   Example 1  Evaluate the following integral.                                                              Solution In this case the substitution   will not work and so we’re going to have to do  something different for this integral.    It would be nice if we could get rid of the square root somehow.  The following substitution  will do that for us.                                                                  Do not worry about where this came from at this point.  As we work the problem you will see  that it works and that if we have a similar type of square root in the problem we can always use a similar substitution.   Before we actually do the substitution however let’s verify the claim that this will allow us to  get rid of the square root.                         To get rid of the square root all we need to do is recall the relationship,                                Using this fact the square root becomes,                                                   Note the presence of the absolute value bars there.  These are important.  Recall that                                                                     There should always be absolute value bars at this stage.  If we knew that    was always positive or always negative we could eliminate the absolute value bars using,                                                              Without limits we won’t be able to determine if   is positive or negative,  however, we will need to eliminate them in order to do the integral.  Therefore, since we are  doing an indefinite integral we will assume that   will be positive and so we  can drop the absolute value bars.  This gives,                                                             So, we were able to eliminate the square root using this substitution.  Let’s now do the  substitution and see what we get.  In doing the substitution don’t forget that we'll also need to  substitute for the dx.  This is easy enough to get from the substitution.                                  Using this substitution the integral becomes,                                        With this substitution we were able to reduce the given integral to an integral involving trig  functions and we saw how to do these problems in the previous section  Let’s finish the  integral.                                                 So, we’ve got an answer for the integral.  Unfortunately the answer isn’t given in x’s as it  should be.  So, we need to write our answer in terms of x.  We can do this with some right  triangle trig.  From our original substitution we have,                                                         This gives the following right triangle. From this we can see that,                                                              We can deal with the   in one of any variety of ways.  From our substitution we can see  that,                                                                 While this is a perfectly acceptable method of dealing with the   we can use any of the  possible six inverse trig functions and since sine and cosine are the two trig functions most  people are familiar with we will usually use the inverse sine or inverse cosine.  In this case  we’ll use the inverse cosine.                                                                So, with all of this the integral becomes,                                   We now have the answer back in terms of x.   Wow!  That was a lot of work.  Most of these won’t take as long to work however.  This first one needed lot’s of explanation since it was the first one.  The remaining examples won’t need quite  as much explanation and so won’t take as long to work.   However, before we move onto more problems let’s first address the issue of definite integrals  and how the process differs in these cases.   Example 2  Evaluate the following integral.                                                             Solution The limits here won’t change the substitution so that will remain the same.                                                                  Using this substitution the square root still reduces down to,                                                            However, unlike the previous example we can’t just drop the absolute value bars.  In this case  we’ve got limits on the integral and so we can use the limits as well as the substitution to  determine the range of   that we’re in.  Once we’ve got that we can determine how to drop  the absolute value bars.   Here’s the limits of  .                                  So, if we are in the range   then   is in the range of   and in this  range of   ’s tangent is positive and so we can just drop the absolute value bars.    Let’s do the substitution.  Note that the work is identical to the previous example and so most  of it is left out.  We’ll pick up at the final integral and then do the substitution.                                             Note that because of the limits we didn’t need to resort to a right triangle to complete the  problem.   Let’s take a look at a different set of limits for this integral.   Example 3  Evaluate the following integral.                                                            Solution Again, the substitution and square root are the same as the first two examples.                                    Let’s next see the limits   for this problem.                               Note that in determining the value of   we used the smallest positive value.  Now for this  range of x’s we have   and in this range of   tangent is  negative and so in this case we can drop the absolute value bars, but will need to add in a  minus sign upon doing so.  In other words,                                                            So, the only change this will make in the integration process is to put a minus sign in front of  the integral.  The integral is then,                                             In the last two examples we saw that we have to be very careful with definite integrals.  We need to make sure that we determine the limits on   and whether or not this will mean that we can  drop the absolute value bars or if we need to add in a minus sign when we drop them.   Before moving on to the next example let’s get the general form for the substitution that we used  in the previous set of examples.     Let’s work a new and different type of example.   Example 4  Evaluate the following integral.                                                              Solution Now, the square root in this problem looks to be (almost) the same as the previous ones so  let’s try the same type of substitution and see if it will work here as well.                                                                    Using this substitution the square root becomes,                                    So using this substitution we will end up with a negative quantity (the tangent squared is  always positive of course) under the square root and this will be trouble.  Using this  substitution will give complex values and we don’t want that.  So, using secant for the  substitution won’t work.   However, the following substitution (and differential) will work.   With this substitution the square root is,                                  We were able to drop the absolute value bars because we are doing an indefinite integral and  so we’ll assume that everything is positive.   The integral is now,                                      In the previous section we saw how to deal with integrals in which the exponent on the secant  was even and since cosecants behave an awful lot like secants we should be able to do  something similar with this.   Here is the integral.                          Now we need to go back to x’s using a right triangle.  Here is the right triangle for this  problem and trig functions for this problem.                                            The integral is then,                                   Here’s the general form for this type of square root.   There is one final case that we need to look at.  The next integral will also contain something that we need to make sure we can deal with.   Example 5  Evaluate the following integral.                                                            Solution First, notice that there really is a square root in this problem even though it isn’t explicitly  written out.  To see the root let’s rewrite things a little.                                          This square root is not in the form we saw in the previous examples.  Here we will use the  substitution for this root.                                       With this substitution the denominator becomes,                                   Now, because we have limits we’ll need to convert them to   so we can determine how to  drop the absolute value bars.                                In this range of   secant is positive and so we can drop the absolute value bars.   Here is the integral,                                      There are several ways to proceed from this point.  Normally with an odd exponent on the  tangent we would strip one of them out and convert to secants.  However, that would require  that we also have a secant in the numerator which we don’t have.  Therefore, it seems like the  best way to do this one would be to convert the integrand to sines and cosines.                                   We can now use the substitution   and we might as well convert the limits as well.                                            The integral is then,                                     The general form for this final type of square root is     We have a couple of final examples to work in this section.  Not all trig substitutions will just  jump right out at us.  Sometimes we need to do a little work on the integrand first to get it into  the correct form and that is the point of the remaining examples.   Example 6  Evaluate the following integral.                                                           Solution In this case the quantity under the root doesn’t obviously fit into any of the cases we looked at  above and in fact isn’t in the any of the forms we saw in the previous examples.  Note however that if we complete the square on the quadratic we can make it look somewhat like the above  integrals.   2 Remember that completing the square requires a coefficient of one in front of the x .  Once we  have that we take half the coefficient of the x, square it, and then add and subtract it to the  quantity.  Here is the completing the square for this problem.                    So, the root becomes,                                                    This looks like a secant substitution except we don’t just have an x that is squared.  That is  okay, it will work the same way.                      Using this substitution the root reduces to,                Note we could drop the absolute value bars since we are doing an indefinite integral.  Here is  the integral.                                 And here is the right triangle for this problem.                                 The integral is then,                Example 7  Evaluate the following integral.                                                                Solution This doesn’t look to be anything like the other problems in this section.  However it is.  To see  this we first need to notice that,                                                                  With this we can use the following substitution.                                           Remember that to compute the differential all we do is differentiate both sides and then tack  on dx or   onto the appropriate side.   With this substitution the square root becomes,                             Again, we can drop the absolute value bars because we are doing an indefinite integral.  Here’s the integral.                     Here is the right triangle for this integral.                                     The integral is then,                                        So, as we’ve seen in the final two examples in this section some integrals that look nothing like  the first few examples can in fact be turned into a trig substitution problem with a little work.   Before leaving this section let’s summarize all three cases in one place.


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