ScholarQuill logoScholarQuillUniversity Notes
  • Notes
  • Past Papers
  • Blogs
  • Todo
Login
ScholarQuill logoScholarQuillUniversity Notes
Login
NotesPast PapersBlogsTodo
More
SubjectsDiscussionCGPA CalculatorGPA CalculatorStudent PortalCourse Outline
About
About usPrivacy PolicyReportContact
Notes
Past Papers
Blogs
Todo
Analytics
    Current Subject
    🧩
    Math Deficiency – II
    MD-002
    Progress0 / 32 topics
    Topics
    1. Complex Numbers2. Arithmetic with Complex Numbers (Add, subtract, multiply and divide complex numbers)3. Trigonometric Polar Form of Complex Numbers4. De Moivre's Theorem and nth Roots5. Recursion6. Sequences and Series7. Sigma Notation8. Arithmetic Series9. Geometric Series (Sum infinite and finite geometric series and categorize geometric series)10. Counting with Permutations and Combinations11. Basic Probability12. Binomial Theorem13. Limit: Notation, Graphs to Find Limits, Tables to Find Limits14. Substitution to Find Limits, Rationalization to Find Limits15. One Sided Limits and Continuity16. Rate of Change: Instantaneous Rate of Change17. Tangent Lines and Rates of Change18. Derivatives: The Derivative Function19. Introduction to Techniques of Differentiation20. The Product and Quotient Rules21. Derivatives of Trigonometric Functions22. The Chain Rule23. Derivatives of Logarithmic Functions24. Derivatives of Exponential and Inverse Trigonometric Functions25. Increase, Decrease, and Concavity26. Relative Extrema, Absolute Maxima and Minima27. Integrals: An Overview of the Area Problem28. Area Under a Curve29. The Indefinite Integral30. Integration by Substitution31. The Definition of Area as a Limit; Sigma Notation32. The Definite Integral
    MD-002›Integration by Substitution
    Math Deficiency – IITopic 30 of 32

    Integration by Substitution

    6 minread
    1,013words
    Intermediatelevel

    Integration by Substitution

    Integration by substitution (also called u-substitution) is a technique used to simplify integrals by transforming them into a more manageable form. It is similar to the chain rule in differentiation but applied in reverse.


    1. The Substitution Rule

    If an integral is of the form:

    ∫f(g(x))g′(x) dx\int f(g(x)) g'(x) \, dx∫f(g(x))g′(x)dx

    we use the substitution:

    u=g(x),so thatdu=g′(x) dxu = g(x), \quad \text{so that} \quad du = g'(x) \, dxu=g(x),so thatdu=g′(x)dx

    This transforms the integral into:

    ∫f(u) du\int f(u) \, du∫f(u)du

    which is often easier to evaluate.


    2. Steps for Integration by Substitution

    1. Choose a substitution: Let u=g(x)u = g(x)u=g(x), where g(x)g(x)g(x) is a function inside the integral.
    2. Differentiate uuu: Compute du=g′(x)dxdu = g'(x) dxdu=g′(x)dx.
    3. Rewrite the integral: Express everything in terms of uuu, replacing dxdxdx with dududu.
    4. Evaluate the new integral: Solve the integral in terms of uuu.
    5. Substitute back xxx: Replace uuu with g(x)g(x)g(x) to return to the original variable.

    3. Examples of Integration by Substitution

    Example 1: ∫2x(x2+1)3 dx\int 2x (x^2 + 1)^3 \, dx∫2x(x2+1)3dx

    Step 1: Choose a substitution
    Let:

    u=x2+1u = x^2 + 1u=x2+1

    Step 2: Differentiate

    du=2x dxdu = 2x \, dxdu=2xdx

    Step 3: Rewrite the integral

    ∫2x(x2+1)3 dx=∫u3 du\int 2x (x^2 + 1)^3 \, dx = \int u^3 \, du∫2x(x2+1)3dx=∫u3du

    Step 4: Integrate

    u44+C\frac{u^4}{4} + C4u4​+C

    Step 5: Substitute back u=x2+1u = x^2 + 1u=x2+1

    (x2+1)44+C\frac{(x^2 + 1)^4}{4} + C4(x2+1)4​+C

    Example 2: ∫sin⁡xcos⁡x dx\int \frac{\sin x}{\cos x} \, dx∫cosxsinx​dx

    Step 1: Choose a substitution
    Let:

    u=cos⁡xu = \cos xu=cosx

    Step 2: Differentiate

    du=−sin⁡x dxdu = -\sin x \, dxdu=−sinxdx

    Step 3: Rewrite the integral

    ∫sin⁡xcos⁡x dx=−∫duu\int \frac{\sin x}{\cos x} \, dx = -\int \frac{du}{u}∫cosxsinx​dx=−∫udu​

    Step 4: Integrate

    −ln⁡∣u∣+C-\ln |u| + C−ln∣u∣+C

    Step 5: Substitute back u=cos⁡xu = \cos xu=cosx

    −ln⁡∣cos⁡x∣+C-\ln |\cos x| + C−ln∣cosx∣+C

    Example 3: Definite Integral ∫01xex2 dx\int_{0}^{1} x e^{x^2} \, dx∫01​xex2dx

    Step 1: Choose a substitution
    Let:

    u=x2u = x^2u=x2

    Step 2: Differentiate

    du=2x dx⇒du2=x dxdu = 2x \, dx \quad \Rightarrow \quad \frac{du}{2} = x \, dxdu=2xdx⇒2du​=xdx

    Step 3: Rewrite the integral

    ∫01xex2 dx=12∫01eu du\int_{0}^{1} x e^{x^2} \, dx = \frac{1}{2} \int_{0}^{1} e^u \, du∫01​xex2dx=21​∫01​eudu

    Step 4: Evaluate the integral

    12eu∣01\frac{1}{2} e^u \Big|_{0}^{1}21​eu​01​ =12(e1−e0)= \frac{1}{2} (e^1 - e^0)=21​(e1−e0) =12(e−1)= \frac{1}{2} (e - 1)=21​(e−1)

    4. When to Use Substitution?

    • When you see a composite function f(g(x))f(g(x))f(g(x)).
    • If the integral contains a function and its derivative (or a multiple of it).
    • When standard integration rules do not apply directly.

    5. Applications of Integration by Substitution

    • Physics: Solving motion equations with changing acceleration.
    • Engineering: Finding energy and work done by varying forces.
    • Probability: Computing probability distributions.
    • Economics: Finding marginal changes in cost and revenue.

    Integration by substitution is one of the most important techniques for solving integrals, helping simplify otherwise complex expressions.

    Previous topic 29
    The Indefinite Integral
    Next topic 31
    The Definition of Area as a Limit; Sigma Notation

    Past Papers

    Open this section to load past papers

    Click on Show Past Papers to see past papers.
    On This Page
      Reading Stats
      Est. reading time6 min
      Word count1,013
      Code examples0
      DifficultyIntermediate