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Methods of Integration

Michael Green

Michael Green

8 min read

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Study Guide Overview

This guide covers integration by substitution (u-substitution), including the basic steps, working with indefinite and definite integrals, and handling more complex substitutions. It provides worked examples, practice questions, and a glossary of key terms. Key takeaways and common mistakes are also highlighted.

Table of Contents

  1. Introduction
  2. Integration by Substitution
    1. Basic Steps for uu-Substitution
    2. Worked Example 1
  3. Advanced uu-Substitution
    1. Worked Example 2
  4. Definite Integrals Using uu-Substitution
    1. Worked Example 3
  5. Practice Questions
  6. Glossary
  7. Summary and Key Takeaways

Introduction

In calculus, integration by substitution (also known as uu-substitution) is a method used to find integrals. It simplifies an integral by transforming it into an easier one through a change of variable.

Integration by Substitution

Basic Steps for uu-Substitution

What is integration by substitution?

Substitution simplifies an integral by defining an alternative variable (usually uu) in terms of the original variable (usually xx). The new integral in uu is often easier to solve, and the substitution can be reversed at the end to express the answer in terms of xx.

How do I integrate simple functions using uu-substitution?

In a simple integral involving substitution, you will usually be integrating a composite function (i.e., 'function of a function'). Substitution can be a safer method when integrating by inspection is awkward or difficult to spot.

Steps for uu-Substitution:

  1. Identify the substitution to be used – it will be the secondary (or 'inside') function in a composite function.

    • If the integral involves f(g(x))f(g(x)), let u=g(x)u = g(x).
    • Example: 3xcos(x25),dx\int 3x \cos(x^2 - 5) , dx
      • Let u=x25u = x^2 - 5.
  2. Differentiate the substitution and rearrange.

    • dudx\frac{du}{dx} can be treated like a fraction. Multiply by dxdx to get rid of fractions.
    • Example: u=x25dudx=2xu = x^2 - 5 \Rightarrow \frac{du}{dx} = 2x
      • Then du=2x,dxx,dx=12,dudu = 2x , dx \Rightarrow x , dx = \frac{1}{2} , du.
  3. Replace all parts of the integral.

    • Replace all uu terms, including dxdx.
    • Example: 3xcos(x25),dx=3cos(x25)x,dx\int 3x \cos(x^2 - 5) , dx = 3 \int \cos(x^2 - 5) \cdot x , dx
      • So 3xcos(x25),dx=3cos(u)12,du=32cos(u),du\int 3x \cos(x^2 - 5) , dx = 3 \int \cos(u) \cdot \frac{1}{2} , du = \frac{3}{2} \int \cos(u) , du.
  4. Integrate.

    • Example: 32cos(u),du=32sin(u)+C\frac{3}{2} \int \cos(u) , du = \frac{3}{2} \sin(u) + C
      • Don't forget the constant of integration.
  5. Substitute xx back in.

    • Replace uu everywhere with the equivalent expression for xx.
    • Example: 32sin(u)+C=32sin(x25)+C\frac{3}{2} \sin(u) + C = \frac{3}{2} \sin(x^2 - 5) + C
      • So 3xcos(x25),dx=32sin(x25)+C\int 3x \cos(x^2 - 5) , dx = \frac{3}{2} \sin(x^2 - 5) + C.

Worked Example 1

Find the indefinite integral x212x36x+7,dx\int \frac{x^2 - 1}{2x^3 - 6x + 7} , dx.

Solution:

  1. Identify the substitution:

    • Let u=2x36x+7u = 2x^3 - 6x + 7.
  2. Differentiate and rearrange:

    • dudx=6x26du=6(x21),dx(x21),dx=16,du\frac{du}{dx} = 6x^2 - 6 \Rightarrow du = 6(x^2 - 1) , dx \Rightarrow (x^2 - 1) , dx = \frac{1}{6} , du.
  3. Replace all parts of the integral:

    • x212x36x+7,dx=1u16,du=161u,du\int \frac{x^2 - 1}{2x^3 - 6x + 7} , dx = \int \frac{1}{u} \cdot \frac{1}{6} , du = \frac{1}{6} \int \frac{1}{u} , du.
  4. Integrate:

    • 161u,du=16lnu+C\frac{1}{6} \int \frac{1}{u} , du = \frac{1}{6} \ln|u| + C.
  5. Substitute xx back in:

    • x212x36x+7,dx=16ln2x36x+7+C\int \frac{x^2 - 1}{2x^3 - 6x + 7} , dx = \frac{1}{6} \ln|2x^3 - 6x + 7| + C.

Advanced uu-Substitution

The procedure here is exactly the same as for integrating simpler functions, but the substitution to use may not be as obvious.

How do I integrate more complicated functions using uu-substitution?

  • Practice more complex questions to improve your integration by substitution skills.
    • Example: xx4,dx\int x \sqrt{x - 4} , dx
      • Identify the substitution: Let u=x4u = x - 4.
      • Differentiate and rearrange: dudx=1du=dx\frac{du}{dx} = 1 \Rightarrow du = dx.
      • Replace all parts of the integral: u=x4x=u+4u = x - 4 \Rightarrow x = u + 4.
        • So xx4,dx=(u+4)u,du\int x \sqrt{x - 4} , dx = \int (u + 4) \sqrt{u} , du.
      • Integrate: (u+4)u,du=(u+4)u12,du=(u32+4u12),du=25u52+83u32+C\int (u + 4) \sqrt{u} , du = \int (u + 4) u^{\frac{1}{2}} , du = \int (u^{\frac{3}{2}} + 4u^{\frac{1}{2}}) , du = \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}} + C.
      • Substitute xx back in: xx4,dx=25(x4)52+83(x4)32+C\int x \sqrt{x - 4} , dx = \frac{2}{5} (x - 4)^{\frac{5}{2}} + \frac{8}{3} (x - 4)^{\frac{3}{2}} + C.

Worked Example 2

Find the indefinite integral dx169x2\int \frac{dx}{\sqrt{16 - 9x^2}}.

Solution:

  1. Identify the substitution:

    • Recall the standard integral dx1x2=arcsin(x)+C\int \frac{dx}{\sqrt{1 - x^2}} = \arcsin(x) + C.
    • Rearrange the integral: dx169x2=dx41916x2=14dx1916x2\int \frac{dx}{\sqrt{16 - 9x^2}} = \int \frac{dx}{4\sqrt{1 - \frac{9}{16}x^2}} = \frac{1}{4} \int \frac{dx}{\sqrt{1 - \frac{9}{16}x^2}}.
  2. Substitute:

    • Let u=34xu2=916x2u = \frac{3}{4}x \Rightarrow u^2 = \frac{9}{16}x^2.
  3. Differentiate and rearrange:

    • dudx=34dx=43,du\frac{du}{dx} = \frac{3}{4} \Rightarrow dx = \frac{4}{3} , du.
  4. Replace all parts of the integral:

    • 14dx1916x2=1443,du1u2=13du1u2\frac{1}{4} \int \frac{dx}{\sqrt{1 - \frac{9}{16}x^2}} = \frac{1}{4} \int \frac{\frac{4}{3} , du}{\sqrt{1 - u^2}} = \frac{1}{3} \int \frac{du}{\sqrt{1 - u^2}}.
  5. Integrate:

    • 13du1u2=13arcsin(u)+C\frac{1}{3} \int \frac{du}{\sqrt{1 - u^2}} = \frac{1}{3} \arcsin(u) + C.
  6. Substitute xx back in:

    • dx169x2=13arcsin(34x)+C\int \frac{dx}{\sqrt{16 - 9x^2}} = \frac{1}{3} \arcsin\left(\frac{3}{4}x\right) + C.

Definite Integrals Using uu-Substitution

How do I evaluate definite integrals using uu-substitution?

Definite integrals can also be solved using uu-substitution. You just need to rewrite the integration limits in terms of uu as well.

Example: 48xx4,dx\int_4^8 x \sqrt{x - 4} , dx

  1. Substitution:

    • Let u=x4u = x - 4.
  2. Differentiate and rearrange:

    • dudx=1du=dx\frac{du}{dx} = 1 \Rightarrow du = dx.
  3. Change the integral limits:

    • When x=4x = 4, u=44=0u = 4 - 4 = 0.
    • When x=8x = 8, u=84=4u = 8 - 4 = 4.
  4. Replace all parts of the integral:

    • 48xx4,dx=04(u+4)u,du\int_4^8 x \sqrt{x - 4} , dx = \int_0^4 (u + 4) \sqrt{u} , du.
  5. Integrate and evaluate the definite integral using the uu values:

    • 04(u+4)u,du=[25u52+83u32]04=2532+838(250+830)=51215\int_0^4 (u + 4) \sqrt{u} , du = \left[ \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}} \right]_0^4 = \frac{2}{5} \cdot 32 + \frac{8}{3} \cdot 8 - \left( \frac{2}{5} \cdot 0 + \frac{8}{3} \cdot 0 \right) = \frac{512}{15}.

Worked Example 3

Evaluate the definite integral 122x+33x2+9x5,dx\int_1^2 \frac{2x + 3}{3x^2 + 9x - 5} , dx.

Solution:

  1. Identify the substitution:

    • Let u=3x2+9x5u = 3x^2 + 9x - 5.
  2. Differentiate and rearrange:

    • dudx=6x+9du=3(2x+3),dx(2x+3),dx=13,du\frac{du}{dx} = 6x + 9 \Rightarrow du = 3(2x + 3) , dx \Rightarrow (2x + 3) , dx = \frac{1}{3} , du.
  3. Find the integration limits in terms of uu:

    • When x=1x = 1, u=3(1)2+9(1)5=7u = 3(1)^2 + 9(1) - 5 = 7.
    • When x=2x = 2, u=3(2)2+9(2)5=25u = 3(2)^2 + 9(2) - 5 = 25.
  4. Replace all parts of the integral, including the integration limits:

    • 122x+33x2+9x5,dx=7251u13,du=137251u,du\int_1^2 \frac{2x + 3}{3x^2 + 9x - 5} , dx = \int_7^{25} \frac{1}{u} \cdot \frac{1}{3} , du = \frac{1}{3} \int_7^{25} \frac{1}{u} , du.
  5. Integrate and evaluate the definite integral:

    • 137251u,du=13[lnu]725=13(ln25ln7)\frac{1}{3} \int_7^{25} \frac{1}{u} , du = \frac{1}{3} \left[ \ln u \right]_7^{25} = \frac{1}{3} (\ln 25 - \ln 7).
  6. Use laws of logarithms to simplify the final answer:

    • 122x+33x2+9x5,dx=13ln(257)\int_1^2 \frac{2x + 3}{3x^2 + 9x - 5} , dx = \frac{1}{3} \ln\left(\frac{25}{7}\right).

Practice Questions

Practice Question
  1. Find the indefinite integral xex2,dx\int x e^{x^2} , dx.

  2. Evaluate the definite integral 012x(x2+1)2,dx\int_0^1 \frac{2x}{(x^2 + 1)^2} , dx.

  3. Find the indefinite integral sin(lnx)x,dx\int \frac{\sin(\ln x)}{x} , dx.

  4. Evaluate the definite integral 143x+2(x2+2x+1),dx\int_1^4 \frac{3x + 2}{(x^2 + 2x + 1)} , dx.

  5. Find the indefinite integral x3x2+1,dx\int \frac{x^3}{x^2 + 1} , dx.

Glossary

  • Composite Function: A function composed of two functions such that the output of one function becomes the input of the other.
  • Constant of Integration: An arbitrary constant added to the function after integration, denoted by CC.
  • Definite Integral: An integral with upper and lower limits, which gives the area under the curve between those limits.
  • Indefinite Integral: An integral without limits, representing a family of functions.
  • Substitution: A method used to simplify an integral by introducing a new variable.

Summary and Key Takeaways

Key Points:

  • Substitution is a powerful method for simplifying integrals, especially for composite functions.
  • Identify the inner function in a composite function to make the correct substitution.
  • Differentiate and rearrange the substitution to replace all parts of the integral.
  • Integrate the simplified integral and always remember to substitute back the original variable.

Key Takeaways:

  • Practice identifying appropriate substitutions in various integrals.
  • Remember to change the limits of integration when evaluating definite integrals.
  • Regular practice with uu-substitution will improve your problem-solving skills in calculus.

Exam Tips:

Exam Tip
  1. Always double-check your substitution and differentiation steps.
  2. Ensure all parts of the integral are replaced correctly before integrating.
  3. Don't forget to add the constant of integration for indefinite integrals.
  4. For definite integrals, remember to adjust the integration limits according to the substitution.

Common Mistakes:

Common Mistake
  1. Forgetting to change the limits of integration when evaluating definite integrals.
  2. Not substituting back the original variable after integrating.
  3. Incorrectly differentiating the substitution.
  4. Omitting the constant of integration for indefinite integrals.

Question 1 of 9

For the integral (2x+1)3dx\int (2x+1)^3 dx, what is the appropriate substitution and the resulting dudu?

u=2x+1u = 2x+1, du=dxdu = dx

u=(2x+1)3u = (2x+1)^3, du=3(2x+1)2dxdu = 3(2x+1)^2 dx

u=2x+1u = 2x+1, du=2dxdu = 2dx

u=x3u = x^3, du=3x2dxdu = 3x^2 dx