Exhaustive Calculus Notes: Integration by Substitution and Definite Integrals

Foundations of Integration by Substitution

  • Core Concept:

    • Integration by substitution extends simple antiderivative rules from the form ∫ex dx\int e^x\,dx to ∫eu du\int e^{u}\,du, where u=g(x)u = g(x) represents a differentiable function of xx.
    • The formula structure remains identical to simple integration rules, but variable transformation is required.
    • If a selected substitution does not simplify the integral, an alternative substitution should be attempted.

  • Methodology & Mental Checks:

    • Identify an inner expression u=g(x)u = g(x) whose derivative du=g′(x) dxdu = g'(x)\,dx appears elsewhere in the integrand, up to a constant multiplier.
    • Constants can be adjusted easily outside the integral using the Constant Multiplier Rule.
    • Regular practice makes appropriate substitution choices immediate.

Standard Algebraic Substitution Example and Verification

  • Evaluating Integral Example 1 (Section 4.5, Page 423, Example 3):

    • Problem Statement: Evaluate ∫x1−4x2 dx\int \frac{x}{\sqrt{1 - 4x^2}}\,dx
    • Strategic Analysis:
      • Let the expression inside the radical be u=1−4x2u = 1 - 4x^2
      • Differentiating yields du=−8x dxdu = -8x\,dx
      • The numerator contains x dxx\,dx, which differs from dudu only by the constant factor −8-8
    • Step-by-Step Solution:
      • Set u=1−4x2u = 1 - 4x^2
      • Calculate the differential: du=−8x dxdu = -8x\,dx
      • Isolate x dxx\,dx: x dx=−18 dux\,dx = -\frac{1}{8}\,du
      • Substitute uu and dxdx into the original integral:             ∫1u(−18 du)\int \frac{1}{\sqrt{u}} \left(-\frac{1}{8}\,du\right)
      • Pull the constant multiplier outside the integral:             −18∫1u du-\frac{1}{8} \int \frac{1}{\sqrt{u}}\,du
      • Rewrite the radical using exponent rules:             −18∫u−1/2 du-\frac{1}{8} \int u^{-1/2}\,du
      • Apply the Power Rule for Integration (∫un du=un+1n+1+C\int u^n\,du = \frac{u^{n+1}}{n+1} + C for n≠−1n \neq -1):             −18(u−1/2+1−1/2+1)+C=−18(u1/21/2)+C-\frac{1}{8} \left( \frac{u^{-1/2 + 1}}{-1/2 + 1} \right) + C = -\frac{1}{8} \left( \frac{u^{1/2}}{1/2} \right) + C
      • Simplify coefficients:             −18×2u1/2+C=−14u1/2+C-\frac{1}{8} \times 2 u^{1/2} + C = -\frac{1}{4} u^{1/2} + C
      • Substitute back the original variable xx:             −141−4x2+C-\frac{1}{4} \sqrt{1 - 4x^2} + C

  • Verification via Differentiation:

    • To verify an indefinite integral, differentiate the result to ensure it yields the original integrand:         ddx[−141−4x2+C]\frac{d}{dx} \left[ -\frac{1}{4} \sqrt{1 - 4x^2} + C \right]
    • Rewrite the radical expression in exponential form:         ddx[−14(1−4x2)1/2+C]\frac{d}{dx} \left[ -\frac{1}{4} (1 - 4x^2)^{1/2} + C \right]
    • Apply the Chain Rule (General Power Rule: ddx[un]=nun−1dudx\frac{d}{dx}[u^n] = n u^{n-1} \frac{du}{dx}):         −14×12(1−4x2)1/2−1×ddx(1−4x2)+0-\frac{1}{4} \times \frac{1}{2} (1 - 4x^2)^{1/2 - 1} \times \frac{d}{dx}(1 - 4x^2) + 0−18(1−4x2)−1/2×(−8x)-\frac{1}{8} (1 - 4x^2)^{-1/2} \times (-8x)
    • Simplify constants and powers:         (−18)×(−8x)×1(1−4x2)1/2=x1−4x2\left(-\frac{1}{8}\right) \times (-8x) \times \frac{1}{(1 - 4x^2)^{1/2}} = \frac{x}{\sqrt{1 - 4x^2}}
    • The derivative matches the integrand, confirming the solution is correct.

Advanced Substitution with Algebraic Inversion

  • Evaluating Integral Example 2:
    • Problem Statement: Evaluate ∫x51+x2 dx\int x^5 \sqrt{1 + x^2}\,dx
    • Strategic Analysis:
      • Let u=1+x2u = 1 + x^2
      • Differentiating gives du=2x dx→dx=du2xdu = 2x\,dx \rightarrow dx = \frac{du}{2x}
      • Substituting dxdx eliminates one factor of xx, leaving x4x^4 in the integrand:             ∫x5u(du2x)=12∫x4u du\int x^5 \sqrt{u} \left(\frac{du}{2x}\right) = \frac{1}{2} \int x^4 \sqrt{u}\,du
      • Since remaining factors of xx exist, express x4x^4 in terms of uu
    • Algebraic Transformation:
      • From u=1+x2u = 1 + x^2, solve for x2x^2: x2=u−1x^2 = u - 1
      • Express x4x^4 as (x2)2(x^2)^2: x4=(u−1)2x^4 = (u - 1)^2
    • Step-by-Step Solution:
      • Substitute x4=(u−1)2x^4 = (u - 1)^2 into the integral:             12∫(u−1)2u1/2 du\frac{1}{2} \int (u - 1)^2 u^{1/2}\,du
      • Expand the binomial (u−1)2=u2−2u+1(u - 1)^2 = u^2 - 2u + 1:             12∫(u2−2u+1)u1/2 du\frac{1}{2} \int (u^2 - 2u + 1) u^{1/2}\,du
      • Distribute u1/2u^{1/2} across the polynomial terms:             12∫(u2+1/2−2u1+1/2+u1/2) du\frac{1}{2} \int \left( u^{2 + 1/2} - 2u^{1 + 1/2} + u^{1/2} \right)\,du12∫(u5/2−2u3/2+u1/2) du\frac{1}{2} \int \left( u^{5/2} - 2u^{3/2} + u^{1/2} \right)\,du
      • Integrate term-by-term using the Power Rule:             12(u7/27/2−2u5/25/2+u3/23/2)+C\frac{1}{2} \left( \frac{u^{7/2}}{7/2} - 2 \frac{u^{5/2}}{5/2} + \frac{u^{3/2}}{3/2} \right) + C
      • Simplify coefficients:             12(27u7/2−45u5/2+23u3/2)+C\frac{1}{2} \left( \frac{2}{7} u^{7/2} - \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right) + C17u7/2−25u5/2+13u3/2+C\frac{1}{7} u^{7/2} - \frac{2}{5} u^{5/2} + \frac{1}{3} u^{3/2} + C
      • Substitute u=1+x2u = 1 + x^2 back into the expression and convert to radical notation:             17(1+x2)7−25(1+x2)5+13(1+x2)3+C\frac{1}{7} \sqrt{(1 + x^2)^7} - \frac{2}{5} \sqrt{(1 + x^2)^5} + \frac{1}{3} \sqrt{(1 + x^2)^3} + C

Derivation of Trigonometric Integrals Using Substitution

  • Derivation of Integral of Tangent:

    • Problem Statement: Evaluate ∫tan⁡(x) dx\int \tan(x)\,dx
    • Trigonometric Identity Rewrite:         ∫sin⁡(x)cos⁡(x) dx\int \frac{\sin(x)}{\cos(x)}\,dx
    • Substitution Attempt Analysis:
      • If u=sin⁡(x)u = \sin(x), then du=cos⁡(x) dx→dx=ducos⁡(x)du = \cos(x)\,dx \rightarrow dx = \frac{du}{\cos(x)}, giving ∫ucos⁡2(x) du\int \frac{u}{\cos^2(x)}\,du, which cannot be easily integrated.
      • If u=cos⁡(x)u = \cos(x), then du=−sin⁡(x) dx→dx=−dusin⁡(x)du = -\sin(x)\,dx \rightarrow dx = -\frac{du}{\sin(x)}.
    • Execution:
      • Let u=cos⁡(x)u = \cos(x)
      • du=−sin⁡(x) dx→dx=−dusin⁡(x)du = -\sin(x)\,dx \rightarrow dx = -\frac{du}{\sin(x)}
      • Substitute into the integral:             ∫sin⁡(x)u(−dusin⁡(x))=−∫1u du\int \frac{\sin(x)}{u} \left(-\frac{du}{\sin(x)}\right) = -\int \frac{1}{u}\,du
      • Integrate using the natural logarithm rule:             −ln⁡(∣u∣)+C-\ln(|u|) + C
      • Substitute back u=cos⁡(x)u = \cos(x), yielding:             −ln⁡(∣cos⁡(x)∣)+C-\ln(|\cos(x)|) + C
    • Logarithmic Simplification to Alternative Form:
      • Using the power property of logarithms pln⁡(a)=ln⁡(ap)p \ln(a) = \ln(a^p), move the coefficient −1-1 into the exponent:             −ln⁡(∣cos⁡(x)∣)=ln⁡(∣cos⁡(x)∣−1)=ln⁡(∣1cos⁡(x)∣)-\ln(|\cos(x)|) = \ln\left(|\cos(x)|^{-1}\right) = \ln\left(\left|\frac{1}{\cos(x)}\right|\right)
      • Since sec⁡(x)=1cos⁡(x)\sec(x) = \frac{1}{\cos(x)}:             ∫tan⁡(x) dx=ln⁡(∣sec⁡(x)∣)+C\int \tan(x)\,dx = \ln(|\sec(x)|) + C

  • Derivation of Integral of Cotangent:

    • Formula Derivation:         ∫cot⁡(x) dx=∫cos⁡(x)sin⁡(x) dx\int \cot(x)\,dx = \int \frac{\cos(x)}{\sin(x)}\,dx
    • Let u=sin⁡(x)u = \sin(x), then du=cos⁡(x) dxdu = \cos(x)\,dx
    • The integral becomes:         ∫1u du=ln⁡(∣u∣)+C\int \frac{1}{u}\,du = \ln(|u|) + C
    • Substituting back u=sin⁡(x)u = \sin(x):         ∫cot⁡(x) dx=ln⁡(∣sin⁡(x)∣)+C\int \cot(x)\,dx = \ln(|\sin(x)|) + C

Definite Integrals and Change of Variables

  • Evaluating Definite Integrals with Substitution:

    • Problem Statement: Evaluate ∫1eln⁡(x)x dx\int_1^e \frac{\ln(x)}{x}\,dx
    • Substitution Setup:
      • Let u=ln⁡(x)u = \ln(x)
      • du=1x dx→dx=x dudu = \frac{1}{x}\,dx \rightarrow dx = x\,du
      • Substitute into integral: ∫ux(x du)=∫u du\int \frac{u}{x} (x\,du) = \int u\,du

  • Evaluation Method 1: Reverting to Original Variable xx:

    • Integrate in terms of uu:         ∫u du=u22\int u\,du = \frac{u^2}{2}
    • Replace uu with ln⁡(x)\ln(x):         [(ln⁡(x))22]1e\left[ \frac{(\ln(x))^2}{2} \right]_1^e
    • Apply the Fundamental Theorem of Calculus (F(e)−F(1)F(e) - F(1)):
      • Upper limit (x=ex = e): (ln⁡(e))22=122=12\frac{(\ln(e))^2}{2} = \frac{1^2}{2} = \frac{1}{2}
      • Lower limit (x=1x = 1): (ln⁡(1))22=022=0\frac{(\ln(1))^2}{2} = \frac{0^2}{2} = 0
    • Subtract: 12−0=12\frac{1}{2} - 0 = \frac{1}{2}

  • Evaluation Method 2: Transforming Limits of Integration to uu:

    • Calculate corresponding uu limits using u=ln⁡(x)u = \ln(x):
      • When x=1x = 1 (lower limit): u=ln⁡(1)=0u = \ln(1) = 0
      • When x=ex = e (upper limit): u=ln⁡(e)=1u = \ln(e) = 1
    • Set up new definite integral strictly in terms of uu:         ∫01u du\int_0^1 u\,du
    • Evaluate directly without substituting back to xx:         [u22]01=122−022=12−0=12\left[ \frac{u^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} - 0 = \frac{1}{2}

Additional Algebraic Substitution Problems and Homework Assignments

  • Evaluating Integral Example 3:

    • Problem Statement: Evaluate ∫xx+2 dx\int x \sqrt{x + 2}\,dx
    • Substitution and Algebraic Inversion Setup:
      • Let u=x+2u = x + 2
      • du=dxdu = dx
      • Solve for xx in terms of uu: x=u−2x = u - 2
    • Step-by-Step Solution:
      • Substitute x=u−2x = u - 2, x+2=ux + 2 = u, and dx=dudx = du:             ∫(u−2)u du\int (u - 2) \sqrt{u}\,du
      • Convert radical to rational exponent:             ∫(u−2)u1/2 du\int (u - 2) u^{1/2}\,du
      • Distribute u1/2u^{1/2}:             ∫(u3/2−2u1/2) du\int \left( u^{3/2} - 2u^{1/2} \right)\,du
      • Integrate term-by-term:             u5/25/2−2(u3/23/2)+C\frac{u^{5/2}}{5/2} - 2 \left(\frac{u^{3/2}}{3/2}\right) + C
      • Simplify constant coefficients:             25u5/2−43u3/2+C\frac{2}{5} u^{5/2} - \frac{4}{3} u^{3/2} + C
      • Substitute back u=x+2u = x + 2 and convert to radical form:             25(x+2)5−43(x+2)3+C\frac{2}{5} \sqrt{(x + 2)^5} - \frac{4}{3} \sqrt{(x + 2)^3} + C

  • Homework Assignment:

    • Section: 4.5
    • Assigned Problems: Number 12, Number 18, Number 52, Number 66, Number 74.
    • Due Date: Friday.