Integration By Parts

7 Techniques of Integration

7.1 Integration by Parts

Introduction to Integration by Parts

  • Every differentiation rule has a corresponding integration rule.

    • Example:

    • The Substitution Rule for integration corresponds to the Chain Rule for differentiation.

    • The rule that corresponds to the Product Rule for differentiation is called integration by parts.

Integration by Parts: Indefinite Integrals

  • The integration by parts formula is derived from the Product Rule for differentiation:

    • If ff and gg are differentiable functions, then
      racddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)rac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

  • Rearranging this gives:

    • extLetuamp;=f(x), vamp;=g(x). extThen,theequationbecomes: racddx[uv]amp;=uracdvdx+vracdudx. extThiscanberearrangedas: uvamp;=racdudx<br>ightarrowudv=d(uv)vdu. extThus,theformulaforintegrationbypartsis: extFormula1:  extIntegrate:amp; <br>tR udv=uvvdu.  This is called the formula for integration by parts. extItmaybemoreeasilyrememberedinthefollowingnotation. extLetu=f(x)extandv=g(x). extThenthedifferentialsaredu=f(x)dxextanddv=g(x)dx. extSo,bytheSubstitutionRule,theformulaforintegrationbypartsbecomes: extIntegrate:  udv=uvvdu.    \begin{aligned} ext{Let } u &amp;= f(x), \ v &amp;= g(x). \ ext{Then, the equation becomes:} \ rac{d}{dx}[uv] &amp;= u rac{dv}{dx} + v rac{du}{dx}. \ ext{This can be rearranged as:} \ uv &amp;= rac{du}{dx} <br>ightarrow u dv = d(uv) - v du. \ ext{Thus, the formula for integration by parts is:} \ ext{Formula 1: } \ \ ext{Integrate:} &amp; \ <br>t R \ u dv = uv - \int v du. \ \ \text{This is called the formula for integration by parts.}\ ext{It may be more easily remembered in the following notation.} \ ext{Let } u = f(x) ext{ and } v = g(x). \ ext{Then the differentials are } du = f'(x)dx ext{ and } dv = g'(x)dx.\ ext{So, by the Substitution Rule, the formula for integration by parts becomes:} \ ext{Integrate: } \ \ u dv = uv - \int v du. \ \ \ \ \end{aligned}

Example 1: Finding an Integral Using Integration by Parts

  • Example Statement: Find xextsin(x)dx\int x ext{ sin}(x) \, dx.

  • From the formula, choose:

    • u=x, dv=extsin(x)dxu = x, \ dv = ext{sin}(x) \, dx.

    • Now find dudu and vv:

    • du=dxdu = dx

    • v=extcos(x)v = - ext{cos}(x)

  • Using Formula 1:

    • xextsin(x)dxamp;=xextcos(x)extcos(x)dx amp;=xextcos(x)+extsin(x)+C, extwhereCamp;extistheconstantofintegration.  \begin{aligned} \int x ext{ sin}(x) \, dx &amp;= -x ext{ cos}(x) - \int - ext{ cos}(x) \, dx \ &amp;= -x ext{ cos}(x) + ext{sin}(x) + C, \ ext{ where } C &amp; ext{ is the constant of integration.} \ \ \end{aligned}

  • Verification: It's wise to check the answer by differentiating the result:

    • Differentiate: racddx[xextcos(x)+extsin(x)+C]=xextsin(x)rac{d}{dx}[-x ext{ cos}(x) + ext{sin}(x) + C] = x ext{ sin}(x), confirming the expected result.

Integration by Parts: Definite Integrals

  • When evaluating definite integrals, the integration by parts formula can be adapted:

    •   extIfuamp;=f(x), v=g(x), extthenwealsohave: <em>abudv=uv</em>ababvdu. \begin{aligned} \ \ ext{If } u &amp;= f(x) , \ v = g(x), \ ext{then we also have:} \ \int<em>a^b u \, dv = uv |</em>{a}^{b} - \int_a^b v \, du. \ \end{aligned}

Example 5: Definite Integral Evaluation

  • Goal: Calculate an1(x)dx\int an^{-1}(x) \, dx.

  • Choose the following for the integration:

    • u=an1(x), du=racdx1+x2, dv=dx, v=xu = an^{-1}(x), \ du = rac{dx}{1 + x^2}, \ dv = dx, \ v = x.

  • Using the integration by parts formula gives:

    •  an1(x)dxamp;=xan1(x)x1+x2dx. \begin{aligned} \ \int an^{-1}(x) \, dx &amp;= x an^{-1}(x) - \int \frac{x}{1 + x^2} \, dx. \ \end{aligned}

  • Evaluating the Integral: The above integral can be solved using substitution. Let:

    • t=1+x2t = 1 + x^2

    • Thus, dt=2xdx<br>ightarrowdx=racdt2xdt = 2x \, dx <br>ightarrow dx = rac{dt}{2x}; change limits accordingly for definite integration.

  • The final evaluation leads to simplified forms and checks against initial conditions to find definite integral solutions.

Reduction Formulas

  • Integration by parts is often used to derive reduction formulas, allowing an integral to be expressed in terms of simpler integrals.

  • Example Statement: Prove the reduction formula:
    sinn(x)dx=sinn1(x)(cos(x))+(n1)sinn2(x)dx\int \sin^n(x) \, dx = \sin^{n-1}(x)(-\cos(x)) + (n-1)\int \sin^{n-2}(x) \, dx,
    where n extisanintegern2n \ ext{ is an integer } n \geq 2.

Example 6: Proving the Reduction Formula
  1. Choose Functions: Let:

    • u=extsin(n1)(x), dv=extsin(x)dxu = ext{sin}(n-1)(x), \ dv = ext{sin}(x) \, dx.

  2. Find Derivatives:

    • du=(n1)extsin(n2)(x)extcos(x)dxdu = (n - 1) ext{sin}(n-2)(x) ext{ cos}(x) dx,

    • v=extcos(x)v = - ext{cos}(x).

  3. Using Integration by Parts Gives:

    •  extsinn(x)dxamp;=extcos(x)extsinn1(x)+(n1)sinn2(x)cos2(x)dx.\begin{aligned} \ \int ext{sin}^n(x) dx &amp;= - ext{cos}(x) ext{sin}^{n-1}(x) + \int (n - 1) \text{sin}^{n-2}(x) \text{cos}^2(x) \, dx. \end{aligned}

  4. Substituting and Rearranging leads to:

    • Deriving a final formula by explicitly solving the rearranged integral form per constants,

    • Showing dependence back on the original integral form.