AOS4 Topic 5: Integration by Substitution

Introduction

Integration by substitution, often referred to as the "u-substitution" method, is a fundamental technique in calculus used to simplify and solve complex integrals. It serves as a powerful tool for mathematicians, scientists, engineers, and students alike, allowing them to navigate through intricate mathematical landscapes with confidence and precision. By strategically substituting variables within integrals, integration by substitution transforms seemingly daunting problems into more manageable forms, unlocking the door to a deeper understanding of functions and their behavior.

Definition

Let \( f(x) \) be a continuous function defined on an interval \([a, b]\), and let \( g(x) \) be a differentiable function on \([a, b]\) such that the range of \( g(x) \) is also contained in the domain of \( f(x) \). Then, the integral of \( f(g(x)) \cdot g'(x) \) with respect to \( x \) over the interval \([a, b]\) can be rewritten as the integral of \( f(u) \) with respect to \( u \) over the interval \([c, d]\), where \( u = g(x) \) and \( c = g(a) \) and \( d = g(b) \).

Mathematically, the integration by substitution formula can be expressed as:

\[ \int_{a}^{b} f(g(x)) \cdot g'(x) \, dx = \int_{c}^{d} f(u) \, du \]

where \( u = g(x) \) and \( du = g'(x) \, dx \).

Why we use integration by substitution?

Integration by substitution is a powerful technique used in calculus for simplifying and solving complex integrals. There are several reasons why integration by substitution is commonly employed:

Simplification of Integrals:

Integration by substitution replaces complex expressions in integrals with simpler ones, facilitating easier evaluation.

Resolution of Composite Functions:

By introducing a new variable, integration by substitution helps untangle composite functions within integrals, making it easier to apply standard integration techniques.

Alignment with Derivative Rules:

This method mirrors the chain rule for derivatives, allowing us to reverse the process and simplify integrals accordingly.

Recognition of Patterns:

Integration by substitution aids in identifying integral patterns resembling derivatives of known functions, streamlining evaluation processes.

Versatility and Generalization:

It offers a broad application scope, applicable to various types of integrals without constraints, making it a versatile tool in calculus.

Application in Real-World Problems:

Integration by substitution finds practical use in scientific and engineering fields, where it enables the analysis of real-world phenomena and facilitates problem-solving in physics, engineering, economics, and beyond.

Change of variable rule:

\[ \int f(u) \, \frac{du}{dx} \, dx = \int f(u) \, du \]

Steps to perform substitution method:

Evaluate the integral \(\int 2x e^{x^2} \, dx\).

  1. Identify the Inner Function:
    In this integral, the inner function is \(x^2\), since the derivative of \(x^2\) is \(2x\), which appears in the integrand.
  2. Choose the New Variable:
    Let's choose the inner function \(x^2\) as the new variable. So, we let \(u = x^2\).
  3. Express the Differential Element:
    To find \(du\), we differentiate \(u\) with respect to \(x\):
    \(du = \frac{d}{dx}(x^2) \, dx = 2x \, dx\)
  4. Rewrite the Integral:
    Substitute \(u = x^2\) and \(du = 2x \, dx\) into the integral:
    \(\int e^u 2x \, dx = \int e^u, du \)
  5. Integrate with Respect to \(u\):
    Now, we integrate with respect to \(u\). The integral becomes:
    \( \int e^u \, du = e^u + C\)
  6. Revert to the Original Variable:
    Replace \(u\) with \(x^2\) to get the final result:
    \(e^{x^2} + C\)
  7. Check the Solution:
    Differentiate \(e^{x^2}\) with respect to \(x\) to verify if it gives the original integrand. The derivative of \(e^{x^2}\) is \(2xe^{x^2}\), which matches the original integrand, confirming the correctness of the solution.

Linear Substitution

Antiderivatives of expressions such as:

\((2x + 3)\sqrt{3x - 4}\), \(\frac{2x + 5}{\sqrt{3x - 4}}\), \(\frac{2x + 5}{(x + 2)^2}\), \((2x + 4)(x + 3)^{20}\), \(x^2 \sqrt{3x - 1}\)

can be found using a linear substitution.

Definite integrals by substitution

Definite integrals by substitution involve applying the technique of substitution to evaluate definite integrals with specified limits of integration. By appropriately choosing the substitution, we can often simplify the integral and make it more manageable for evaluation.

Antiderivatives of expressions such as:

\[ \int_{0}^{1} 2x \sqrt{1 - x^2} \, dx \],\[ \int_{0}^{1} e^{2x} \, dx \]

can be found using a definite integral substitution.

Use of Trigonometric Identities for Integration

Products of sines and cosines:

Integrals of the form \( \int \sin^m x \cos^n x \, dx \), where \( m \) and \( n \) are non-negative integers, can be considered in the following three cases:

Case A: The power of sine is odd

If \( m \) is odd, write \( m = 2k + 1 \). Then

\[ \sin^{2k+1} x = (\sin^2 x)^k \sin x = (1 - \cos^2 x)^k \sin x \]

and the substitution \( u = \cos x \) can now be made.

Case B:The power of cosine is odd

If \( m \) is even and \( n \) is odd, write \( n = 2k + 1 \). Then

\[ \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1 - \sin^2 x)^k \cos x \]

and the substitution \( u = \sin x \) can now be made.

Case C: Both powers are even

If both \( m \) and \( n \) are even, then the identity

\[ \sin^2 x = \frac{1}{2}(1 - \cos(2x)), \quad \cos^2 x = \frac{1}{2}(1 + \cos(2x)), \quad \text{or} \quad \sin(2x) = 2 \sin x \cos x \]

can be used.

Also note that \( \int \sec^2(kx) \, dx = \frac{1}{k} \tan(kx) + c \). The identity \( 1 + \tan^2 x = \sec^2 x \) is used in the following example.

Product-to-Sum Identities

2 cos x cos y = cos(x − y) + cos(x + y)

2 sin x sin y = cos(x − y) − cos(x + y)

2 sin x cos y = sin(x + y) + sin(x − y)

These identities enable us to determine further integrals involving the circular functions.

Further Substitution

\[ \int f(u) \, \frac{du}{dx} \, dx = \int f(u) \, du \]

If we interchange the variables \(x\) and \(u\), then we can write this as follows:

\[ \int f(x) \, dx = \int f(x) \, \frac{dx}{du} \, du \]

Note: For this substitution to work, the function that we substitute for \(x\) must be one-to-one.

Example 1

Evaluate the integral \( \int 2x e^{x^2} \, dx \).

Solution:
  • Identify the Inner Function:
    In this integral, the inner function is \( x^2 \), since the derivative of \( x^2 \) is \( 2x \), which appears in the integrand.
  • Choose the New Variable:
    Let's choose the inner function \( x^2 \) as the new variable. So, we let \( u = x^2 \).
  • Express the Differential Element:
    To find \( du \), we differentiate \( u \) with respect to \( x \):
    \( du = \frac{d}{dx}(x^2) \, dx = 2x \, dx \)

  • Rewrite the Integral:
    Substitute \( u = x^2 \) and \( du = 2x \, dx \) into the integral:
    \( \int 2x e^u \, du \)
  • Integrate with Respect to \( u \):
    Now, we integrate with respect to \( u \). The integral becomes:
    \( 2 \int e^u \, du = 2e^u + C \)
  • Revert to the Original Variable:
    Replace \( u \) with \( x^2 \) to get the final result:
    \( 2e^{x^2} + C \)
  • Check the Solution:
    Differentiate \( 2e^{x^2} \) with respect to \( x \) to verify if it gives the original integrand. The derivative of \( 2e^{x^2} \) is \( 2xe^{x^2} \), which matches the original integrand, confirming the correctness of the solution.

    • Example 2

    Find the Antiderivative of \(5x^2(x^3 - 1)^{\frac{1}{2}}\).

    Solution:
    Let \(u = x^3 - 1\).
    Then \(f(u) = u^{\frac{1}{2}}\) and \(\frac{du}{dx} = 3x^2\).
    So, \(\int 5x^2(x^3 - 1)^{\frac{1}{2}} \, dx = \frac{5}{3} \int (x^3 - 1)^{\frac{1}{2}} \cdot 3x^2 \, dx\)
    Now, \(\int (x^3 - 1)^{\frac{1}{2}} \cdot 3x^2 \, dx = \frac{5}{3} \int u^{\frac{1}{2}} \frac{du}{dx} \, dx\)

    After integration, we get: \(\frac{5}{3} \int u^{\frac{1}{2}} \, du = \frac{5}{3} \left[ \frac{2}{3} u^{\frac{3}{2}} \right] + c\)
    Substituting back \(u = x^3 - 1\), we have: \(\frac{10}{9} u^{\frac{3}{2}} + c = \frac{10}{9} (x^3 - 1)^{\frac{3}{2}} + c\)
    So, the antiderivative is: \(\frac{10}{9} (x^3 - 1)^{\frac{3}{2}} + c\)

    Example 3

    Evaluate the integral \( \int \frac{-2x}{\sqrt{4-x^2}} \, dx \).

    Solution:
    To evaluate the integral \(\int \frac{-2x}{\sqrt{4-x^2}} \, dx\), we use the substitution method.
    Let \(u = 4 - x^2\), then \(\frac{du}{dx} = -2x\) and \(dx = -\frac{du}{2x}\).
    Substituting \(u = 4 - x^2\) and \(dx = -\frac{du}{2x}\) into the integral:

    \[ \int \frac{-2x}{\sqrt{4-x^2}} \, dx = \int \frac{-2x}{\sqrt{u}} \left( -\frac{du}{2x} \right) \]

    Simplifying the expression:

    \[ = \int \frac{du}{\sqrt{u}} \]

    Integrating with respect to \(u\):

    \[ = 2 \sqrt{u} + C \]

    Substituting back \(u = 4 - x^2\):

    \[ = 2 \sqrt{4 - x^2} + C \]

    So, the solution to the integral \(\int \frac{-2x}{\sqrt{4-x^2}} \, dx\) is \(2 \sqrt{4 - x^2} + C\).

    Example 4

    Find the antiderivative of \( \frac{2}{x^2 + 2x + 6} \)

    Solution:
    Completing the square gives:
    \( x^2 + 2x + 6 = x^2 + 2x + 1 + 5 = (x + 1)^2 + 5 \)
    Therefore, \( \int \frac{2}{x^2 + 2x + 6} \, dx = \int \frac{2}{(x + 1)^2 + 5} \, dx \)
    Let \( u = x + 1 \). Then \( \frac{du}{dx} = 1 \) and hence
    \( \int \frac{2}{(x + 1)^2 + 5} \, dx = \int \frac{2}{u^2 + 5} \, du \)

    After simplification, we get:
    \( \frac{2}{\sqrt{5}} \int \frac{\sqrt{5}}{u^2 + 5} \, du \)
    Using the arctangent integral formula, we have:
    \( \frac{2}{\sqrt{5}} \tan^{-1} \left( \frac{u}{\sqrt{5}} \right) + c \)
    Substituting back \( u = x + 1 \), we get:
    \( \frac{2}{\sqrt{5}} \tan^{-1} \left( \frac{x + 1}{\sqrt{5}} \right) + c \)

    Example 5

    Find the Antiderivative of \( \frac{3}{\sqrt{9 - 4x - x^2}} \) .

    Solution:
    Completing the square gives:
    \( 9 - 4x - x^2 = -(x^2 + 4x - 9) = -(x + 2)^2 - 13 \)
    Therefore, \( \int \frac{3}{\sqrt{9 - 4x - x^2}} \, dx = \int \frac{3}{\sqrt{13 - (x + 2)^2}} \, dx \)
    Let \( u = x + 2 \). Then \( \frac{du}{dx} = 1 \) and hence
    \( \int \frac{3}{\sqrt{13 - (x + 2)^2}} \, dx = \int \frac{3}{\sqrt{13 - u^2}} \, du \)

    After simplification, we get:
    \( 3 \sin^{-1} \left( \frac{u}{\sqrt{13}} \right) + c \)
    Substituting back \( u = x + 2 \), we get:
    \( 3 \sin^{-1} \left( \frac{x + 2}{\sqrt{13}} \right) + c \)

    Example 6

    Evaluate the integral:\(\int \frac{2x + 1}{(1 - 2x)^2} \, dx\)

    Solution:
    Given the integral:

    \[ \int \frac{2x + 1}{(1 - 2x)^2} \, dx \]

    We can solve it using the substitution method:
    1. Let \( u = 1 - 2x \).
    2. Then, \( \frac{du}{dx} = -2 \) and \( 2x = 1 - u \).
    3. Substitute \( u = 1 - 2x \) and \( 2x = 1 - u \) into the integral:

    \[ \int \frac{2x + 1}{(1 - 2x)^2} \, dx = \int \frac{2 - u}{u^2} \cdot (-2) \, dx \]

    Simplify the expression:

    \[ = -\frac{1}{2} \int \frac{2 - u}{u^2} \, du \]

    \[ = -\frac{1}{2} \int \frac{2u - 2}{u^2} \, du \]

    \[ = -\frac{1}{2} \int \left( \frac{2}{u} - \frac{2}{u^2} \right) \, du \]

    \[ = -\frac{1}{2} \left( -2 \ln|u| - \frac{2}{u} \right) + C \]

    \[ = \ln|u| - \frac{1}{u} + C \]

    \[ = \ln|1 - 2x| - \frac{1}{1 - 2x} + C \]

    \[ = \frac{1}{1 - 2x} + \frac{1}{2} \ln|1 - 2x| + C \]

    So, the antiderivative of \( \frac{2x + 1}{(1 - 2x)^2} \) is \( \frac{1}{1 - 2x} + \frac{1}{2} \ln|1 - 2x| + C \).

    Example 7

    Evaluate the Integral by linear substitution: \( \int x^2\sqrt{3x - 1} \, dx \)

    Solution:
    Given the integral:

    \[ \int x^2\sqrt{3x - 1} \, dx \]

    We can solve it using the substitution method:
    1. Let \( u = 3x - 1 \).
    2. Then, \( \frac{du}{dx} = 3 \).
    3. We have \( x = \frac{u + 1}{3} \) and so \( x^2 = \frac{(u + 1)^2}{9} \).
    4. Substitute \( u = 3x - 1 \) and \( x^2 = \frac{(u + 1)^2}{9} \) into the integral:

    \[ \int x^2\sqrt{3x - 1} \, dx = \int \frac{(u + 1)^2}{9}\sqrt{u} \, du \]

    Simplify the expression:

    \[ = \frac{1}{27} \int \left( (u + 1)^2 \right) u^{\frac{1}{2}} \cdot 3 \, du \]

    \[ = \frac{1}{27} \int (u^{\frac{5}{2}} + 2u^{\frac{3}{2}} + u^{\frac{1}{2}}) \, du \]

    \[ = \frac{1}{27} \left( \frac{2}{7}u^{\frac{7}{2}} + \frac{4}{5}u^{\frac{5}{2}} + \frac{2}{3}u^{\frac{3}{2}} \right) + C \]

    \[ = \frac{2}{2835} u^{\frac{3}{2}} \left( 135u^2 + 36u + 8 \right) + C \]

    \[ = \frac{2}{2835} (3x - 1)^{\frac{3}{2}} \left( 135x^2 + 36x + 8 \right) + C \]

    So, the antiderivative of \( \int x^2\sqrt{3x - 1} \, dx \) is:

    \[ \frac{2}{2835}(3x - 1)^{\frac{3}{2}} \left( 135x^2 + 36x + 8 \right) + C \]

    Example 8

    Evaluate \( \int_{0}^{4} 3x \sqrt{x^2 + 9} \, dx \)

    Solution:
    Let \( u = x^2 + 9 \). Then \( \frac{du}{dx} = 2x \) and so

    \[ \int 3x \sqrt{x^2 + 9} \, dx = \frac{3}{2} \int \sqrt{x^2 + 9} \cdot 2x \, dx \]

    \[ = \frac{3}{2} \int u^{\frac{1}{2}} \, du \]

    \[ = \frac{3}{2} \left( \frac{2}{3} u^{\frac{3}{2}} \right) + c \]

    \[ = u^{\frac{3}{2}} + c \]

    \[ = (x^2 + 9)^{\frac{3}{2}} + c \]

    \[ \int_{0}^{4} 3x \sqrt{x^2 + 9} \, dx = \left( (x^2 + 9)^{\frac{3}{2}} \right) \Bigg|_{0}^{4} \]

    Therefore,

    \[ = \left( (4^2 + 9)^{\frac{3}{2}} \right) - \left( (0^2 + 9)^{\frac{3}{2}} \right) \]

    \[ = \left( 25^{\frac{3}{2}} \right) - \left( 9^{\frac{3}{2}} \right) \]

    \[ = 125 - 27 = 98 \]

    Example 9

    Evaluate the integral:\( \int_{-2}^{-1} \frac{e^x}{1 - e^x} \, dx \)

    Solution:
    To solve the integral \( \int_{-2}^{-1} \frac{e^x}{1 - e^x} \, dx \), we use the substitution method.
    Let \( u = 1 - e^x \). Then \( du = -e^x \, dx \), which implies \( dx = -\frac{1}{e^x} \, du \). Now, we need to express the integral in terms of \( u \).
    When \( x = -2 \), \( u = 1 - e^{-2} \), and when \( x = -1 \), \( u = 1 - e^{-1} \).
    The integral becomes:
    \[ \int_{1 - e^{-2}}^{1 - e^{-1}} \frac{-e^{-x}}{u} \, du \]

    Now, we can integrate with respect to \( u \).
    \[ = \int_{1 - e^{-2}}^{1 - e^{-1}} -\frac{1}{u} \, du \]
    \[ = -\ln|u| \Bigg|_{1 - e^{-2}}^{1 - e^{-1}} \]
    \[ = -\ln(1 - e^{-1}) + \ln(1 - e^{-2}) \]
    \[ = \ln\left(\frac{1 - e^{-2}}{1 - e^{-1}}\right) \]
    \[ \approx \ln\left(\frac{1 - \frac{1}{e^2}}{1 - \frac{1}{e}}\right) \]
    \[ \approx \ln\left(\frac{\frac{e^2 - 1}{e^2}}{\frac{e - 1}{e}}\right) \]
    \[ = \ln\left(\frac{e^2 - 1}{e - 1}\right) \]

    Example 10

    Find \( \int \cos^4(x) \, dx \).

    Solution:
    Given \( \cos^4(x) = \left(\frac{\cos(2x) + 1}{2}\right)^2 = \frac{1}{4} (\cos(2x) + 1)^2 = \frac{1}{4} (\cos^2(2x) + 2\cos(2x) + 1) \)
    As \( \cos(4x) = 2\cos^2(2x) - 1 \), this gives:
    \[ \cos^4(x) = \frac{1}{4} \left( \cos(4x) + 1 + 2\cos(2x) + 1 \right) = \frac{1}{8} \cos(4x) + \frac{1}{2} \cos(2x) + \frac{3}{8} \]

    So, the integral becomes:
    \[ \int \cos^4(x) \, dx = \int \left( \frac{1}{8} \cos(4x) + \frac{1}{2} \cos(2x) + \frac{3}{8} \right) \, dx = \frac{1}{32} \sin(4x) + \frac{1}{4} \sin(2x) + \frac{3}{8}x + C \]
    This is the solution expressed in terms of trigonometric functions and a constant term.

    Example 11

    Find \( \int \sin^3(x) \cos^2(x) \, dx \)

    Solution:
    We have \( \int \sin^3(x) \cos^2(x) \, dx \).
    Breaking down \( \sin^3(x) \cos^2(x) \) as \( \sin(x) (\sin^2(x) \cos^2(x)) \), we get:
    \[ \int \sin x (1 - \cos^2(x)) \cos^2(x) \, dx \]

    Now, let \( u = \cos(x) \). Then \( du/dx = -\sin(x) \).

    We obtain:
    \[ \int \sin^3(x) \cos^2(x) \, dx = -\int (-\sin(x))(1 - u^2)(u^2) \, dx \]
    \[ = -\int (1 - u^2)u^2 \, du \]
    \[ = -\int (u^2 - u^4) \, du \]
    \[ = -\left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C \]
    \[ = -\left( \frac{\cos^3(x)}{3} - \frac{\cos^5(x)}{5} \right) + C \]
    Therefore, \( \int \sin^3(x) \cos^2(x) \, dx = \frac{\cos^5(x)}{5} - \frac{\cos^3(x)}{3} + C \).

    Example 12

    Integrate: \( \int \sin(3x) \sin(2x) \, dx \)

    Solution:
    We need to find \( \int \sin(3x) \sin(2x) \, dx \).
    Using the identity \( \sin(A) \sin(B) = \frac{1}{2}[\cos((A - B)) - \cos((A + B))] \), we have:
    \[ \int \sin(3x) \sin(2x) \, dx = \frac{1}{2} \int [\cos((3x - 2x)) - \cos((3x + 2x))] \, dx \]
    \[ = \frac{1}{2} \int [\cos(x) - \cos(5x)] \, dx \]

    Integrating each term separately, we get:
    \[ = \frac{1}{2} [\sin(x) - \frac{1}{10}\sin(5x)] + C \]
    Therefore, \( \int \sin(3x) \sin(2x) \, dx = \frac{1}{2} [\sin(x) - \frac{1}{10}\sin(5x)] + C \).

    Example 13

    Evaluate \( \int \frac{1}{(x^2 + 1)^2} \, dx \) by using the substitution \( x = \tan(u) \), where \( -\frac{\pi}{2} < u < \frac{\pi}{2} \).

    Solution
    Let \( x = \tan(u) \). Then \( \frac{dx}{du} = \sec^2(u) \).
    We substitute into \( \int f(x) \, dx = \int f(x) \, \frac{dx}{du} \, du \).
    \( \int \frac{1}{(x^2 + 1)^2} \, dx = \int \frac{1}{(\tan^2(u) + 1)^2} \cdot \sec^2(u) \, du \)
    \( = \int \frac{1}{\sec^4(u)} \cdot \sec^2(u) \, du \) since \( 1 + \tan^2(u) = \sec^2(u) \)
    \( = \int \cos^2(u) \, du \)

    \( = \frac{1}{2} \int (\cos(2u) + 1) \, du \) since \( \cos^2(u) = \frac{1 + \cos(2u)}{2} \)
    \( = \frac{1}{2} \left( \frac{1}{2} \sin(2u) + u \right) + c \)
    Since \( x = \tan(u) \), we have \( \sin(u) = \frac{x}{\sqrt{x^2 + 1}} \) and \( \cos(u) = \frac{1}{\sqrt{x^2 + 1}} \).
    \( \int \frac{1}{(x^2 + 1)^2} \, dx = \frac{1}{2} \cdot \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} + \frac{1}{2} \arctan(x) + c \)
    Hemce,\( \int \frac{1}{(x^2 + 1)^2} \, dx = \frac{x}{2\sqrt{x^2 + 1}} + \frac{1}{2} \arctan(x) + c \)

    Exercise 1

    Exercise 2

    Exercise 3

    Exercise 4

    Exercise 5

    Exercise 6

    Exercise 7

    Exercise 8

    Exercise 9

    Exercise 10

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