Calculus Study Guide: Integration of Exponential, Rational, Inverse Trigonometric, and Hyperbolic Functions

Course Logistics and Assessment Structure

  • Class Delivery and Schedule:

    • The course is currently in Week 4, operating online, with planned transitions to face-to-face/blended modes in upcoming weeks.
    • Key scheduled blended face-to-face dates include September 19, October 10, and November 14.
    • The Prelim Exam period is scheduled for Week 6 (September 7 to September 12).
    • Final exams are scheduled for November 26 to November 28.

  • Assessment Details:

    • Quiz 1 is scheduled for Week 5 (face-to-face if allowed) covering all material from the start of the term through Exponential Formulas.
    • Quiz 2 covers Trigonometric through Hyperbolic formulas and will occur after Prelim exams (administered via Blackboard online or as a take-home exam).
    • Students are permitted to bring index cards containing formula references (derivative, antiderivative, trigonometric, and hyperbolic identities) to all quizzes.
    • Late submissions for assignments are accepted on Blackboard subject to point deductions. Partial points/consideration are granted for showing complete solution steps.

Review of Assignment Problems

  • Assignment Problem 1: Indefinite Integration of Exponential and Hyperbolic Functions

    • Given Problem: Evaluate the integral \n\int \left( 3^{\cosh(x)} + e^{\cosh(x)} \right)^2 \sinh(x)\,dx\n
    • Substitution Step: Let u=cosh⁡(x)u = \cosh(x). The derivative is du=sinh⁡(x) dxdu = \sinh(x)\,dx.
    • Rewritten Integral: \n\int (3^u + e^u)^2\,du\n
    • Expanding the Square: \n\int \left( 3^{2u} + 2(3^u e^u) + e^{2u} \right)du = \int \left( 3^{2u} + 2(3e)^u + e^{2u} \right)du\n
    • Integration Formula: \n\int a^u\,du = \frac{a^u}{\ln(a)} + C\n
    • Term-by-Term Integration:
    • For 32u3^{2u}: Let v=2uv = 2u, dv=2 du  ⟹  du=12 dvdv = 2\,du \implies du = \frac{1}{2}\,dv. The integral gives \n\frac{3^{2u}}{2\ln(3)}\n
    • For 2(3e)u2(3e)^u: Constant factor remains 22, giving \n\frac{2(3e)^u}{\ln(3e)} = \frac{6 e^u}{\ln(3e)}\n
    • For e2ue^{2u}: Let v=2uv = 2u, yielding \n\frac{e^{2u}}{2}\n
    • Final Answer (substituting back u=cosh⁡(x)u = \cosh(x)):     \n    \frac{3^{2\cosh(x)}}{2\ln(3)} + \frac{6 e^{\cosh(x)}}{\ln(3e)} + \frac{e^{2\cosh(x)}}{2} + C\n    

  • Assignment Problem 2: Rational Exponential Integration

    • Given Problem: Evaluate the integral \n\int \frac{e^{2x} + 2e^x}{e^x + 1}\,dx\n
    • Method 1: Algebraic Manipulation / Factoring Numerator
    • Factor numerator: e2x+2ex=ex(ex+2)=ex(ex+1+1)=ex(ex+1)+exe^{2x} + 2e^x = e^x(e^x + 2) = e^x(e^x + 1 + 1) = e^x(e^x + 1) + e^x
    • Rewrite integrand: \n    \frac{e^x(e^x + 1) + e^x}{e^x + 1} = e^x + \frac{e^x}{e^x + 1}\n    
    • Method 2: Polynomial Long Division
    • Since the degree of the numerator exponent e2xe^{2x} is higher than the denominator exponent exe^x, divide (e2x+2ex)(e^{2x} + 2e^x) by (ex+1)(e^x + 1).
    • Division process: (e2x+2ex)÷(ex+1)=ex(e^{2x} + 2e^x) \div (e^x + 1) = e^x with a remainder of exe^x.
    • Integrand form: \n    e^x + \frac{e^x}{e^x + 1}\n    
    • Integration Step:     \n    \int e^x\,dx + \int \frac{e^x}{e^x + 1}\,dx\n    
    • For the second integral, let u=ex+1u = e^x + 1, then du=ex dxdu = e^x\,dx.
    • Integral becomes \n    \int \frac{du}{u} = \ln|u|\n    
    • Final Answer:     \n    e^x + \ln(e^x + 1) + C\n    

  • True or False Conceptual Questions:

    • Statement: If \n  \int f(x)\,dx = \frac{1}{2}\sin(2x) + C\n  , then f(x)=12cos⁡(2x)f(x) = \frac{1}{2}\cos(2x).
    • Evaluation: False.
    • Proof/Correction: To find f(x)f(x), differentiate the right-hand side:     \n    f(x) = \frac{d}{dx}\left[ \frac{1}{2}\sin(2x) \right] = \frac{1}{2} \cdot \cos(2x) \cdot 2 = \cos(2x)\n         Therefore, f(x)=cos⁡(2x)f(x) = \cos(2x), making the original statement false.

  • Test 3 Problem 1: Trigonometric Substitution

    • Given Problem: Evaluate \n  \int \frac{\sec^2(\cos(4t))}{\csc(4t)}\,dt\n  
    • Identity Simplification: Since \n  \frac{1}{\csc(4t)} = \sin(4t)\n  , rewrite as:     \n    \int \sec^2(\cos(4t)) \sin(4t)\,dt\n    
    • Substitution Step: Let u=cos⁡(4t)u = \cos(4t). Using the chain rule, du=−4sin⁡(4t) dt  ⟹  sin⁡(4t) dt=−14 dudu = -4\sin(4t)\,dt \implies \sin(4t)\,dt = -\frac{1}{4}\,du
    • Rewritten Integral: \n  -\frac{1}{4} \int \sec^2(u)\,du\n  
    • Integration Step: The integral of \n  \sec^2(u)\n   is \n  \tan(u)\n  
    • Final Answer:     \n    -\frac{1}{4}\tan(\cos(4t)) + C\n    

Inverse Trigonometric Functions

  • Standard Derivatives of Inverse Trigonometric Functions:

    • \n  \frac{d}{dx}[\arcsin(u)] = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}\n  
    • \n  \frac{d}{dx}[\arccos(u)] = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}\n  
    • \n  \frac{d}{dx}[\arctan(u)] = \frac{1}{1 + u^2} \cdot \frac{du}{dx}\n  
    • \n  \frac{d}{dx}[\text{arccot}(u)] = -\frac{1}{1 + u^2} \cdot \frac{du}{dx}\n  
    • \n  \frac{d}{dx}[\text{arcsec}(u)] = \frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx}\n  
    • \n  \frac{d}{dx}[\text{arccsc}(u)] = -\frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx}\n  

  • Standard Integration Formulas for Inverse Trigonometric Forms:

    • Form 1 (Arcsine):     \n    \int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left( \frac{u}{a} \right) + C\n    
    • Form 2 (Arctangent):     \n    \int \frac{du}{a^2 + u^2} = \frac{1}{a} \arctan\left( \frac{u}{a} \right) + C\n    
    • Form 3 (Arcsecant):     \n    \int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{a} \text{arcsec}\left( \frac{u}{a} \right) + C\n    

  • Worked Examples for Inverse Trigonometric Integrals:

    • Example 1: Evaluate \n  \int \frac{dx}{\sqrt{9 - x^2}}\n  

    • Identify terms: u2=x2  ⟹  u=x,du=dxu^2 = x^2 \implies u = x, du = dx; a2=9  ⟹  a=3a^2 = 9 \implies a = 3

    • Apply Arcsine formula:       \n      \arcsin\left( \frac{x}{3} \right) + C\n      

    • Example 2: Evaluate \n  \int \frac{d\theta}{16 + 4\theta^2}\n  

    • Identify terms: u2=4θ2  ⟹  u=2θ,du=2 dθ  ⟹  dθ=12 duu^2 = 4\theta^2 \implies u = 2\theta, du = 2\,d\theta \implies d\theta = \frac{1}{2}\,du; a2=16  ⟹  a=4a^2 = 16 \implies a = 4

    • Substitution:       \n      \frac{1}{2} \int \frac{du}{16 + u^2} = \frac{1}{2} \cdot \frac{1}{4} \arctan\left( \frac{2\theta}{4} \right) + C\n      

    • Simplification:       \n      \frac{1}{8} \arctan\left( \frac{\theta}{2} \right) + C\n      

    • Example 3: Evaluate \n  \int \frac{dy}{5y\sqrt{25y^2 - 121}}\n  

    • Identify terms: u2=25y2  ⟹  u=5y,du=5 dy  ⟹  dy=15 duu^2 = 25y^2 \implies u = 5y, du = 5\,dy \implies dy = \frac{1}{5}\,du; a2=121  ⟹  a=11a^2 = 121 \implies a = 11

    • Rewrite in Arcsecant form: Note that denominator contains 5y=u5y = u       \n      \frac{1}{5} \int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{5} \cdot \frac{1}{11} \text{arcsec}\left( \frac{5y}{11} \right) + C\n      

    • Simplification:       \n      \frac{1}{55} \text{arcsec}\left( \frac{y}{11} \right) + C\n      

    • Example 4: Evaluate \n  \int \frac{e^{2m}}{\sqrt{36 - e^{4m}}}\,dm\n  

    • Identify terms: u2=e4m  ⟹  u=e2m,du=2e2m dm  ⟹  e2m dm=12 duu^2 = e^{4m} \implies u = e^{2m}, du = 2e^{2m}\,dm \implies e^{2m}\,dm = \frac{1}{2}\,du; a2=36  ⟹  a=6a^2 = 36 \implies a = 6

    • Apply Arcsine formula:       \n      \frac{1}{2} \int \frac{du}{\sqrt{a^2 - u^2}} = \frac{1}{2} \arcsin\left( \frac{e^{2m}}{6} \right) + C\n      

    • Example 5: Evaluate \n  \int \frac{dy}{\sqrt{20 + 8y - y^2}}\n  

    • Data Manipulation (Completing the Square under radical):

      • Rearrange polynomial: 20−(y2−8y)20 - (y^2 - 8y)
      • Determine completing square constant: (−82)2=16\left( \frac{-8}{2} \right)^2 = 16
      • Add and subtract 16: 20−(y2−8y+16)+16=36−(y−4)220 - (y^2 - 8y + 16) + 16 = 36 - (y - 4)^2

    • Identify terms: u2=(y−4)2  ⟹  u=y−4,du=dyu^2 = (y - 4)^2 \implies u = y - 4, du = dy; a2=36  ⟹  a=6a^2 = 36 \implies a = 6

    • Rewrite integral:       \n      \int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left( \frac{y - 4}{6} \right) + C\n      

    • Example 6: Evaluate \n  \int \frac{3x - 2}{x^2 + 6x + 10}\,dx\n  

    • Completing the Square in Denominator:

      • x2+6x+10=(x2+6x+9)+1=(x+3)2+1x^2 + 6x + 10 = (x^2 + 6x + 9) + 1 = (x + 3)^2 + 1

    • Numerator Decomposition:

      • Express 3x−23x - 2 in terms of (x+3)(x + 3): 3x−2=3(x+3)−9−2=3(x+3)−113x - 2 = 3(x + 3) - 9 - 2 = 3(x + 3) - 11

    • Split Integral into Two Parts:       \n      \int \frac{3(x + 3)}{(x + 3)^2 + 1}\,dx - \int \frac{11}{(x + 3)^2 + 1}\,dx\n      

    • First Integral Part (Logarithmic Rule):

      • Let v=(x+3)2+1=x2+6x+10v = (x + 3)^2 + 1 = x^2 + 6x + 10
      • dv=2(x+3) dx  ⟹  (x+3) dx=12 dvdv = 2(x + 3)\,dx \implies (x + 3)\,dx = \frac{1}{2}\,dv
      • \n      3 \cdot \frac{1}{2} \int \frac{dv}{v} = \frac{3}{2}\ln(x^2 + 6x + 10)\n      

    • Second Integral Part (Arctangent Rule):

      • Let u=x+3u = x + 3, du=dxdu = dx, a=1a = 1
      • \n      -11 \int \frac{du}{u^2 + 1^2} = -11 \arctan(x + 3)\n      

    • Final Combined Result:       \n      \frac{3}{2}\ln(x^2 + 6x + 10) - 11\arctan(x + 3) + C\n      

Hyperbolic Functions

  • Standard Integration Formulas for Hyperbolic Functions:

    • \n  \int \sinh(u)\,du = \cosh(u) + C\n  
    • \n  \int \cosh(u)\,du = \sinh(u) + C\n  
    • \n  \int \text{sech}^2(u)\,du = \tanh(u) + C\n  
    • \n  \int \text{csch}^2(u)\,du = -\coth(u) + C\n  
    • \n  \int \text{sech}(u)\tanh(u)\,du = -\text{sech}(u) + C\n  
    • \n  \int \text{csch}(u)\coth(u)\,du = -\text{csch}(u) + C\n  
    • \n  \int \tanh(u)\,du = \ln|\cosh(u)| + C\n  
    • \n  \int \coth(u)\,du = \ln|\sinh(u)| + C\n  
    • \n  \int \text{sech}(u)\,du = 2\arctan(e^u) + C\n   or \n  \arctan(\sinh(u)) + C\n  
    • \n  \int \text{csch}(u)\,du = \ln|\coth(u) - \text{csch}(u)| + C\n   or \n  \ln\left| \tanh\left(\frac{1}{2}u\right) \right| + C\n  

  • Worked Examples for Hyperbolic Integrals:

    • Example 1: Evaluate \n  \int \cosh(5x)\,dx\n  

    • Substitution: Let u=5xu = 5x, du=5 dx  ⟹  dx=15 dudu = 5\,dx \implies dx = \frac{1}{5}\,du

    • Calculation:       \n      \frac{1}{5} \int \cosh(u)\,du = \frac{1}{5}\sinh(5x) + C\n      

    • Example 2: Evaluate \n  \int \frac{\tanh(\ln(\cos(x)))}{\cot(x)}\,dx\n  

    • Identity Simplification: Since \n    \frac{1}{\cot(x)} = \tan(x)\n    , rewrite as:       \n      \int \tanh(\ln(\cos(x))) \tan(x)\,dx\n      

    • Substitution: Let u=ln⁡(cos⁡(x))u = \ln(\cos(x))

    • Derivative: Using chain rule, du=1cos⁡(x)⋅(−sin⁡(x)) dx=−tan⁡(x) dx  ⟹  tan⁡(x) dx=−dudu = \frac{1}{\cos(x)} \cdot (-\sin(x))\,dx = -\tan(x)\,dx \implies \tan(x)\,dx = -du

    • Calculation:       \n      -\int \tanh(u)\,du = -\ln|\cosh(\ln(\cos(x)))| + C\n      

    • Example 3: Evaluate \n  \int \frac{dt}{t \sinh^2\left( \ln\left( \frac{2}{3t} \right) \right)}\n  

    • Identity Simplification: Using \n    \frac{1}{\sinh^2(\theta)} = \text{csch}^2(\theta)\n    , rewrite as:       \n      \int \text{csch}^2\left( \ln\left( \frac{2}{3t} \right) \right) \frac{dt}{t}\n      

    • Substitution: Let u=ln⁡(23t)u = \ln\left( \frac{2}{3t} \right)

    • Derivative calculation:       \n      du = \frac{1}{\frac{2}{3t}} \cdot \left( -\frac{2}{3t^2} \right) dt = \frac{3t}{2} \cdot \left( -\frac{2}{3t^2} \right) dt = -\frac{dt}{t} \implies \frac{dt}{t} = -du\n      

    • Calculation:       \n      -\int \text{csch}^2(u)\,du = -(-\coth(u)) + C = \coth\left( \ln\left( \frac{2}{3t} \right) \right) + C\n      

    • Example 4: Evaluate \n  \int \frac{e^{2\theta} \sinh(\text{arccot}(e^{2\theta}))}{1 + e^{4\theta}}\,d\theta\n  

    • Substitution: Let u=arccot(e2θ)u = \text{arccot}(e^{2\theta})

    • Derivative calculation:       \n      du = -\frac{1}{1 + (e^{2\theta})^2} \cdot (2e^{2\theta})\,d\theta = -\frac{2e^{2\theta}}{1 + e^{4\theta}}\,d\theta \implies \frac{e^{2\theta}\,d\theta}{1 + e^{4\theta}} = -\frac{1}{2}\,du\n      

    • Calculation:       \n      -\frac{1}{2} \int \sinh(u)\,du = -\frac{1}{2}\cosh(\text{arccot}(e^{2\theta})) + C\n      

    • Example 5: Evaluate \n  \int \frac{x^3 \text{csch}^2\left( \sqrt{x^4 + 1} \right)}{\sqrt{x^4 + 1}}\,dx\n  

    • Substitution: Let u=x4+1=(x4+1)1/2u = \sqrt{x^4 + 1} = (x^4 + 1)^{1/2}

    • Derivative calculation:       \n      du = \frac{1}{2}(x^4 + 1)^{-1/2} \cdot (4x^3)\,dx = \frac{2x^3}{\sqrt{x^4 + 1}}\,dx \implies \frac{x^3\,dx}{\sqrt{x^4 + 1}} = \frac{1}{2}\,du\n      

    • Calculation:       \n      \frac{1}{2} \int \text{csch}^2(u)\,du = \frac{1}{2}(-\coth(u)) + C = -\frac{1}{2}\coth\left( \sqrt{x^4 + 1} \right) + C\n      

Student Questions and Interactive Discussions

  • Question regarding denominator simplification in Example 3 of Inverse Trigonometric Functions:
    • Student Inquiry: Whether the constant factor 55 in denominator term 5y5y cancels directly out of the inverse trigonometric angle.
    • Clarification: The factor 55 forms part of the variable term u=5yu = 5y. When applying the arcsecant formula \n  \frac{1}{a}\text{arcsec}\left(\frac{u}{a}\right)\n  , the substitution u=5yu = 5y gives \n  \text{arcsec}\left(\frac{5y}{11}\right)\n  . Angle values are distinct from outer multiplicative constants and cannot be arbitrarily canceled outside function arguments.