Math 305                     Methods of Integration
       The following are a list of integration formulas that you should know. Even when you go through the
       different methods, you reduce the integrals to one of these:
           Z          n+1                        Z
              n     x                                 2
             x dx = n+1+cn6=−1                      sec xdx = tanx+c
           Z dx                                   Z   2
              x =ln|x|+c                            csc xdx = −cotx+c
           Z x       x                           Z
             e dx = e +c                           secxtanxdx = secx+c
           Z sinxdx =−cosx+c                     Z cscxcotxdx = −cscx+c
           Z cosxdx = sinx+c
       The following are nice to know (make your life easier) integrals:
           Z tanxdx =−ln|cosx|+c                Z cscxdx = −ln|cscx+cotx|+c
           Z cotxdx =ln|sinx|+c                  Z secxdx =ln|secx+tanx|+c
       If a is constant, a 6=0:
           Z ax      1 ax          Z           1                Z            1
             e  dx = ae  +c          cosaxdx = a sinax+c          sinaxdx = −a cosax+c
       HintI:Ifyouthinkyouhaveanantiderivative,justdifferentiateandseeifyouendupwiththeintegrand!
       Hint II: The line of attack you should take when trying to integrate should be to try these methods in
       this order. You will always need to have paper and pencil, the correct method doesn’t usually bounce
       off the page and hit you in the face!
          Math 305                                                                                             MI-Page 2
          Method 1. Straight forward integration. The integral can be reduced to one of the first set of integrals.
                        Z                        Z                              4          2
                             2                        3      2                x       3   x
          Example A.       (x +1)(x−3)dx =          (x −3x +x−3)dx= 4 −x + 2 −3x+c
                        Z          2        Z                               Z
          Example B.       1−sin xdx =         (1 −sinx)(1+sinx)dx =          (1 −sinx)dx = x+cosx+c
                            1+sinx                    1+sinx
          Problems:
                  Z x        −x     2
               1.    e (1−e      sec x)dx
                  Z x3−2x+4
               2.         x       dx
               3. Z cosxtanxdx
          Method 2. Substitution. A simple substitution reduces the integral to one of the first set of integrals.
          In fact, the second set came about by using substitution.
          Example C. Z tanxdx = Z sinxdx. Let w = cosx then dw = −sinxdx so that the integral becomes
          Z                              cosx
            −dw =−ln|w|+c=−ln|cosx|+c.
               w
                       Z      2                                                                   Z     dw      1
                             x                 2                                                       w          w
          Example D.      xe dx. Let w = x then dw =2xdx so that the integral becomes                e   2 = 2e +c=
          1   2
           ex +c.
          2
          Hint III: Things to look for: if the integrand involves
                                                 f(x)                  1           n
                                                e    ; trig (f(x)); f(x); (f(x))
          Let w = f(x): This is not an exclusive list!
          Hint IV: When substitution is complete, make sure no x’s appear in the integrand.
        Math 305                                                                            MI-Page 3
        Problems:
            4. Z   cosx dx. Let w=1+sinx:
                 1+sinx
               Z    2          3    2                   3    2
            5.   (3x +x)cos(2x +x +4)dx. Let w =2x +x +4:
            6. Z (lnx)3dx.  Let w =lnx.
                    x
               Z    2  tanx
            7.   sec xe    dx
               Z  √
            8.   x x+2dx
            9. Z  2 x+1 dx
                 x +2x−5
        Method 3. Integration by parts. This method undoes the product rule
                                             Z udv = uv−Z vdu
                   Z                                                       2
        Example E.    xlnxdx: Let u =lnx; dv = xdx then du = dx and v = x .
                                                               x          2
                               Z                 2   Z   2       2        2
                                 xlnxdx =(lnx)x −       x dx = x lnx− x +c
                                                 2      2 x      2       4
        Example F. Z xsinxdx: Let u = x; dv = sinxdx: Then dv = dx and v = −cosx.
                           Z xsinxdx = x(−cosx)−Z −cosxdx =−xcosx+sinx+c
         Math 305                                                                                    MI-Page 4
         Hint V: When choosing u and dv make sure dv is something that can be integrated. Also, the whole
         integrand should be taken up with u and dv.
         Hint VI: Method should be used when integrand involves
                                                      ax         ax                           (2n+1)
                          (poly)trig                 e  (sinbx) e   cosbx                   sec      x
                                                                                               (2n+1)
                       (poly) lnx                   poly(inverse trig fact)                 csc      x
             This is not an exclusive list!
         Problems:
             10. Z lnxdx.    Let u =lnx; dv = dx:
                 Z    2                               2
             11.   (x +2x−1)cos3xdx. Let u=x +2x−1;dv=cos3xdx:
                                             (Need to use integration by parts twice.)
                 Z         2x                            2x
             12.   (x +3)e dx. Let u=x+3;dv=e dx:
                 Z   2
             13.    x lnxdx
                 Z     −1
             14.    tan   xdx
                 Z x
             15.    e sinxdx
         Method 4. Trigonometric integrals. Integrands only involve trigonometric functions (not inverse trig
         functions!). Remember that certain trig functions “go together.”
                                                     sinθ and cosθ
                                                    tanθ and secθ
                                                     cotθ and cotθ
         If you have mixed trig functions convert everything to sinθ and cosθ. Another helpful identity is
           2        2                                    2           2             2        2
         sin θ+cos θ =1: From here you can derive tan θ+1 = sec θ and 1+cot θ = csc θ: You should also
         have the half angle formulas in your repertoire.
                                                    2     1
                                                  sin θ =2(1−cos2θ)
                                                    2     1
                                                 cos θ =2(1+cos2θ)