jagomart
digital resources
picture1_Methods Of Integration Pdf 90158 | Methint


 166x       Filetype PDF       File size 0.17 MB       Source: www.math.cmu.edu


File: Methods Of Integration Pdf 90158 | Methint
methods of integration william gunther june 15 2011 in this we will go over some of the techniques of integration and when to apply them 1 simple rules so remember ...

icon picture PDF Filetype PDF | Posted on 15 Sep 2022 | 3 years ago
Partial capture of text on file.
                                           Methods of Integration
                                                 William Gunther
                                                   June 15, 2011
                 In this we will go over some of the techniques of integration, and when to apply them.
              1   Simple Rules
              So, remember that integration is the inverse operation to differentation. Thuse we get a few rules for free:
                   Sum/Difference R(f(x)±g(x))dx=R f(x)dx±R g(x)dx
                   Scalar Multiplication R cf(x) dx = c·R f(x) dx for c ∈ R
                                R        xn+1
                   Product Rule   xn dx =     +Cfor n6=−1
                                         n+1
              The above allows us to integrate any polynomials and roots. The only think we don’t yet know how to
              integrate is R 1 dx. Luckily, we know d ln(x) = 1. From this, and other knowledge we know about
                          x                     dx       x
              derivatives, we know:
                   Trig
                   R sin(x) dx = −cos(x)+C
                   R cos(x) dx = sin(x) +C
                   R sec2(x) dx = tan(x)+C
                   R sec(x)tan(x) dx = sec(x)+C
                   Exponentials
                   R x       x
                     e dx=e +C
                   R 1 dx = ln|x|+C.
                     x
                 !!EXAMPLES!!
              2   u-substitution
              Notice, if f(x) and g(x) are functions, then the chain rule says
                                              d (f(g(x))) = f′(g(x))·g′(x)
                                              dx
              So, we know:                   Z
                                               f′(g(x)) · g′(x) dx = f(g(x))
                 Writing this out in a better way, we get let u = g(x). Then du = g′(x) dx, meaning we can trade a
              g′(x) dx for a du and substitute u for g(x) in the integral. The goal is to eliminate all occurrences of x in
              the integral, and then your entire integral is in terms of u, and is simplier.
                                                         1
                Example 1. Let us solve the integral         Z
                                                               sin(2x) dx
                We do this by doing the substitution u = 2x. Then du = 2 dx. Thus we can trade a 2 dx for a du. So we
                write the integral in the following way:
                                                   Z sin(2x) dx = 1 Z sin(2x)(2 dx)
                                                                  2
                Then:                                Z                  Z
                                                   1   sin(2x)(2 dx) = 1   sin(u) du
                                                   2                   2
                Doing the integration:               Z
                                                    1   sin(u) du = 1(−cos(u))+C
                                                    2              2
                As the problem was given in terms of x, we want the answer in terms of x. So we substitute 2x for u.
                                                   1(−cos(u))+C =−cos(2x) +C
                                                   2                      2
                   Wedothe following integrals with less exposition:
                Example 2.                                  Z
                                                                     2
                                                               xcos(x ) dx
                Set u = x2. Then du = 2x dx.
                                                  Z        2       1 Z       2
                                                     xcos(x ) dx = 2    cos(x )2x dx
                                                                 = 1 Z cos(u) du
                                                                   2
                                                                 = 1(sin(u))+C
                                                                   2
                                                                   sin(x2)
                                                                 =    2    +C
                Example 3.                                 Z
                                                              cos(ln(x)) dx
                                                                  x
                Set u = ln(x). Then du = 1 dx.
                                         x
                                                  Z cos(ln(x)) dx = Z cos(ln(x))1 dx
                                                        x           Z           x
                                                                  = cos(u)du
                                                                  =sin(u)+C
                                                                  =sin(ln(x))+C
                Example 4.                                Z
                                                            3cos(x)esin(x) dx
                Let u = sin(x). Then du = cos(x) dx.
                                               Z         sin(x)       Z         sin(x)
                                                 3cos(x)e      dx = 3·Z cos(x)e      dx
                                                                          u
                                                                  =3·    e du
                                                                  =3·eu+C
                                                                  =3esin(x) +C
                                                                  2
              Example 5.                           Z
                                                           10
                                                     x(x+5) dx
              Here, we can solve the integral by expanding. But, expanding the 10th power is rather annoying. So instead:
                 Let u = x+5. Then we get x = u−5, and du = dx. All we have done is a linear transformation. Note,
              in general we can not solve for x when we do a substitution. When the substitution is linear we can.
                                      Z x(x+5)10 dx = Z (u−5)(u)10 du
                                                      Z  11    10
                                                    = (u −5u )du
                                                       12    11
                                                    = u  −5u +C
                                                      12    11
                                                      (x+5)12   5(x+5)11
                                                    =    12   −    11   +C
              3   Integration by Parts
              Recall the product rule:
                                           d [f(x)g(x)] = f′(x)g(x)+f(x)g′(x)
                                          dx
              Moving things around, we see
                                          f′(x)g(x) = d [f(x)g(x)]−f(x)g′(x)
                                                    dx
              Integrating both sides, we see
                                       Z f′(x)g(x) dx = f(x)g(x)−Z f(x)g′(x)dx
              Renaming v = f(x) and u = g(x) we have dv = f′(x) dx and du = g′(x) dx and our formula becomes
                                                Z udv =uv−Z vdu
                 Here, we seperate our integral into two parts: one part we differentiate, and the other we integrate. Then
              we apply the formula, and get a new integral with these new parts (the derivative of the one part and the
              integral of the other).
                 As a strategy, we tend to choose our u (the part we differentiate) so that the new integral is easier to
              integrate. We also need to take care that the dv (the part we integrate) can actually be integrated by us.
              Example 6.                             Z
                                                          x
                                                       x·e dx
              Here, we see that when we take the derivate of x it vanishes completely making our next integral simplier.
                             Z                                                   x
                               x·ex dx                             u=x      dv = e dx
                                                                   du = dx v = ex
                                          x  Z x
                                      =xe − e dx
                                          x   x
                                      =xe −e +C
                 Asaheuristic (rule of thumb) we choose logarithms and inverse trigonometric functions to be our u before
              any others since their integrals are hard to calculate and complicated. After those, we like polynomials (or
              really anything algebraic) as those derivatives often get simplier. We rarely want to choose exponentials to
              be our u since integrating an exponential is virtually the same as deriving it.
                 Unless we deviate from this heuristic, the u shall be chosen without exposition:
                                                         3
                   Example 7.                                              Z
                                                                              ln(x) dx
                   Wedothis by parts:
                                       Z ln(x) dx                                                 u=ln(x)       dv = dx
                                                                                                  du = 1 dx     v = x
                                                                  Z                                     x
                                                     =xln(x)−         1xdx
                                                                  Z x
                                                     =xln(x)−         dx
                                                     =xln(x)−x+C
                   Example 8.                                          Z
                                                                         1
                                                                             2       −x
                                                                          (x +1)e        dx
                                                                        0
                   Wedothis by parts:
                          Z 1                                                                                   2               −x
                              (x2 +1)e−x dx                                                              u=x +1 dv=e dx
                           0                                           Z                                 du = 2x dx      v = −e−x
                                                     2          −x            −x
                                               =(x +1)(−e )− (−e )(2x)dx
                                                       2       −x      Z    −x                               u=x         dv = e−x dx
                                               =−(x +1)e          +2 xe dx                                   du = dx v = −e−x
                                                       2       −x           −x    Z      −x
                                               =−(x +1)e          +2(−xe       − (−e )dx)
                                                       2       −x        −x      Z −x
                                               =−(x +1)e          −2xe      +2 e dx
                                                       2       −x        −x       −x
                                               =−(x +1)e          −2xe      −2e      +C
                       Sometimes, we can do a nice subsitution before finishing the problem using parts:
                   Example 9.                                             Z
                                                                                    2
                                                                              3    x
                                                                             x ·3     dx
                                                                                                            b     bln(a)
                   First, we re-write it to have e as the exponential base. Note that in general a = e                   So
                                                              Z    3   x2       Z    3   x2ln(3)
                                                                 x ·3     dx =     x ·e          dx
                   Let u = x2. Then du = 2x dx. So 1 du = x dx.
                                                           2
                                                        Z         2            Z            2
                                                             3   x ln(3)            2     (x )ln(3)
                                                           x ·e          dx = Z (x )·e             (x dx)
                                                                                       uln(3)
                                                                             = u·e           du
                   Now, we do parts.
                                                                                4
The words contained in this file might help you see if this file matches what you are looking for:

...Methods of integration william gunther june in this we will go over some the techniques and when to apply them simple rules so remember that is inverse operation dierentation thuse get a few for free sum dierence r f x g dx scalar multiplication cf c xn product rule cfor n above allows us integrate any polynomials roots only think don t yet know how luckily d ln from other knowledge about derivatives trig sin cos sec tan exponentials e examples u substitution notice if are functions then chain says z writing out better way let du meaning can trade substitute integral goal eliminate all occurrences your entire terms simplier example solve do by doing thus write following as problem was given want answer wedothe integrals with less exposition xcos set esin eu here expanding but th power rather annoying instead have done linear transformation note general not parts recall moving things around see integrating both sides renaming v dv our formula becomes udv uv vdu seperate into two one par...

no reviews yet
Please Login to review.