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Calculus II: Techniques of Integration
Calculus II: Techniques of Integration
Calculus II: Techniques of Integration
> This document covers the fundamental techniques of integration presented in Calculus II. Each module contains theorems, key concepts, and worked examples. Click on an example to view the detailed solution.
Module 1: The Substitution Rule
Theorem: The Substitution Rule. If u=g(x) is a differentiable function whose range is an interval I, and f is continuous on I, then:
∫ f(g(x))g'(x) dx = ∫ f(u) du
Key Concepts:
This technique reverses the Chain Rule for derivatives.
Identify an "inner function" g(x) to define as u.
The derivative of u (i.e., g'(x)dx) must be present in the integral, possibly off by a constant factor.
For definite integrals, convert the limits of integration from x-values to u-values.
Example 1.1: Evaluate ∫ x2√(x3 + 1) dx
1. DEFINE u: Let u be the inner function.
u = x³ + 1
2. FIND du: Differentiate u with respect to x.
du = 3x² dx
3. REWRITE INTEGRAL: Adjust for the constant factor.
(1/3) ∫ √(x³ + 1) * (3x² dx)
4. SUBSTITUTE: Replace all terms with u and du.
(1/3) ∫ √u du = (1/3) ∫ u¹⁄² du
5. INTEGRATE: Apply the power rule for integration.
(1/3) * [ (2/3)u³⁄² ] + C = (2/9)u³⁄² + C
6. BACK-SUBSTITUTE: Convert the result back to x.
(2/9)(x³ + 1)³⁄² + C
Module 2: Integration by Parts
Theorem: Integration by Parts. Derived from the Product Rule for derivatives.
∫ u dv = uv - ∫ v du
Key Concepts:
Used for integrating a product of two functions.
Choose u using the **L.I.P.E.T.** priority list:
**L**ogarithmic, **I**nverse Trig, **P**olynomial, **E**xponential, **T**rigonometric.
The remaining part of the integrand becomes dv.
This process may need to be applied multiple times.
Example 2.1: Evaluate ∫ ln(x) dx
1. CHOOSE u AND dv: Following LIPET, Logarithmic is chosen for u.
u = ln(x) dv = dx
du = (1/x) dx v = x
2. APPLY FORMULA: uv - ∫ v du
x*ln(x) - ∫ x * (1/x) dx
3. SIMPLIFY: The new integral is simpler.
x*ln(x) - ∫ 1 dx
4.…
Calculus II: Techniques of Integration
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<h1>Calculus II: Techniques of Integration</h1>
<p>> This document covers the fundamental techniques of integration presented in Calculus II. Each module contains theorems, key concepts, and worked examples. Click on an example to view the detailed solution.</p>
<!-- MODULE 1: U-SUBSTITUTION -->
<div class="module">
<h2>Module 1: The Substitution Rule</h2>
<div class="theorem-box">
<strong>Theorem: The Substitution Rule.</strong> If <span class="math">u=g(x)</span> is a differentiable function whose range is an interval I, and f is continuous on I, then:
<br><span class="integral">∫</span> <span class="math">f</span>(<span class="math">g</span>(<span class="math">x</span>))<span class="math">g</span>'(<span class="math">x</span>) d<span class="math">x</span> = <span class="integral">∫</span> <span class="math">f</span>(<span class="math">u</span>) d<span class="math">u</span>
</div>
<h3>Key Concepts:</h3>
<ul class="concept-list">
<li>This technique reverses the Chain Rule for derivatives.</li>
<li>Identify an "inner function" <span class="math">g(x)</span> to define as <span class="math">u</span>.</li>
<li>The derivative of <span class="math">u</span> (i.e., <span class="math">g'(x)dx</span>) must be present in the integral, possibly off by a constant factor.</li>
<li>For definite integrals, convert the limits of integration from <span class="math">x</span>-values to <span class="math">u</span>-values.</li>
</ul>
<div class="example" onclick="toggleSolution(this)">Example 1.1: Evaluate <span class="integral">∫</span> <span class="math">x</span><sup>2</sup><span class="math">√</span>(<span class="math">x</span><sup>3</sup> + 1) d<span class="math">x</span></div>
<div class="solution">
<code>
1. DEFINE u: Let u be the inner function.
u = x³ + 1
2. FIND du: Differentiate u with respect to x.
du = 3x² dx
3. REWRITE INTEGRAL: Adjust for the constant factor.
(1/3) ∫ √(x³ + 1) * (3x² dx)
4. SUBSTITUTE: Replace all terms with u and du.
(1/3) ∫ √u du = (1/3) ∫ u¹⁄² du
5. INTEGRATE: Apply the power rule for integration.
(1/3) * [ (2/3)u³⁄² ] + C = (2/9)u³⁄² + C
6. BACK-SUBSTITUTE: Convert the result back to x.
(2/9)(x³ + 1)³⁄² + C
</code>
</div>
</div>
<!-- MODULE 2: INTEGRATION BY PARTS -->
<div class="module">
<h2>Module 2: Integration by Parts</h2>
<div class="theorem-box">
<strong>Theorem: Integration by Parts.</strong> Derived from the Product Rule for derivatives.
<br><span class="integral">∫</span> <span class="math">u</span> d<span class="math">v</span> = <span class="math">uv</span> - <span class="integral">∫</span> <span class="math">v</span> d<span class="math">u</span>
</div>
<h3>Key Concepts:</h3>
<ul class="concept-list">
<li>Used for integrating a product of two functions.</li>
<li>Choose <span class="math">u</span> using the **L.I.P.E.T.** priority list:
<br>**L**ogarithmic, **I**nverse Trig, **P**olynomial, **E**xponential, **T**rigonometric.</li>
<li>The remaining part of the integrand becomes <span class="math">dv</span>.</li>
<li>This process may need to be applied multiple times.</li>
</ul>
<div class="example" onclick="toggleSolution(this)">Example 2.1: Evaluate <span class="integral">∫</span> <span class="math">ln</span>(<span class="math">x</span>) d<span class="math">x</span></div>
<div class="solution">
<code>
1. CHOOSE u AND dv: Following LIPET, Logarithmic is chosen for u.
u = ln(x) dv = dx
du = (1/x) dx v = x
2. APPLY FORMULA: uv - ∫ v du
x*ln(x) - ∫ x * (1/x) dx
3. SIMPLIFY: The new integral is simpler.
x*ln(x) - ∫ 1 dx
4. INTEGRATE:
x*ln(x) - x + C
</code>
</div>
</div>
<!-- MODULE 3: TRIG INTEGRALS -->
<div class="module">
<h2>Module 3: Trigonometric Integrals</h2>
<div class="theorem-box">
<strong>Key Identities:</strong>
<br>sin<sup>2</sup><span class="math">x</span> + cos<sup>2</sup><span class="math">x</span> = 1
<br>tan<sup>2</sup><span class="math">x</span> + 1 = sec<sup>2</sup><span class="math">x</span>
<br>sin<sup>2</sup><span class="math">x</span> = (1 - cos(2<span class="math">x</span>))/2
<br>cos<sup>2</sup><span class="math">x</span> = (1 + cos(2<span class="math">x</span>))/2
</div>
<h3>Strategy:</h3>
<ul class="concept-list">
<li>For <span class="integral">∫</span>sin<sup>m</sup><span class="math">x</span>cos<sup>n</sup><span class="math">x</span> dx:</li>
<li>If the power of <span class="math">cos(x)</span> is ODD, save one <span class="math">cos(x)</span> factor and convert the rest to <span class="math">sin(x)</span>. Use <span class="math">u=sin(x)</span>.</li>
<li>If the power of <span class="math">sin(x)</span> is ODD, save one <span class="math">sin(x)</span> factor and convert the rest to <span class="math">cos(x)</span>. Use <span class="math">u=cos(x)</span>.</li>
<li>If BOTH powers are EVEN, use the half-angle identities to reduce the powers.</li>
</ul>
<div class="example" onclick="toggleSolution(this)">Example 3.1: Evaluate <span class="integral">∫</span> cos<sup>4</sup>(<span class="math">x</span>) d<span class="math">x</span></div>
<div class="solution">
<code>
1. STRATEGY: The power is even, so use the half-angle identity.
∫ (cos²(x))² dx = ∫ [ (1 + cos(2x))/2 ]² dx
2. EXPAND THE INTEGRAND:
∫ (1/4) * (1 + 2cos(2x) + cos²(2x)) dx
3. APPLY IDENTITY AGAIN: Another even power requires another reduction.
cos²(2x) = (1 + cos(4x))/2
4. SUBSTITUTE AND SIMPLIFY:
∫ (1/4) * [1 + 2cos(2x) + (1 + cos(4x))/2] dx
∫ (1/4) * [3/2 + 2cos(2x) + (1/2)cos(4x)] dx
∫ [3/8 + (1/2)cos(2x) + (1/8)cos(4x)] dx
5. INTEGRATE TERM BY TERM:
(3/8)x + (1/2)*(1/2)sin(2x) + (1/8)*(1/4)sin(4x) + C
(3/8)x + (1/4)sin(2x) + (1/32)sin(4x) + C
</code>
</div>
</div>
<!-- MODULE 4: TRIG SUBSTITUTION -->
<div class="module">
<h2>Module 4: Trigonometric Substitution</h2>
<div class="theorem-box">
<strong>Method:</strong> For integrands with quadratic expressions in radicals, transform the integral into the trigonometric domain using a reference triangle.
</div>
<h3>Strategy:</h3>
<ul class="concept-list">
<li>For <span class="math">√(a² - x²)</span>, let <span class="math">x = a*sin(θ)</span>, so <span class="math">dx = a*cos(θ)dθ</span>.</li>
<li>For <span class="math">√(a² + x²)</span>, let <span class="math">x = a*tan(θ)</span>, so <span class="math">dx = a*sec²(θ)dθ</span>.</li>
<li>For <span class="math">√(x² - a²)</span>, let <span class="math">x = a*sec(θ)</span>, so <span class="math">dx = a*sec(θ)tan(θ)dθ</span>.</li>
<li>After integrating, use the reference triangle to convert the result back to x.</li>
</ul>
<div class="example" onclick="toggleSolution(this)">Example 4.1: Evaluate <span class="integral">∫</span> <span class="frac"><span class="numer">dx</span><span class="denom"><span class="math">x</span><sup>2</sup><span class="math">√</span>(9 - <span class="math">x</span><sup>2</sup>)</span></span></div>
<div class="solution">
<code>
1. IDENTIFY FORM: The expression matches √(a² - x²) with a=3.
Let x = 3sin(θ), which means dx = 3cos(θ)dθ.
2. SUBSTITUTE INTO THE INTEGRAL:
∫ 3cos(θ)dθ / [ (3sinθ)² * √(9 - 9sin²θ) ]
3. SIMPLIFY THE RADICAL:
√(9(1-sin²θ)) = √(9cos²θ) = 3cos(θ)
4. SIMPLIFY THE FULL INTEGRAL:
∫ 3cos(θ)dθ / [ 9sin²θ * 3cos(θ) ]
∫ 1 / (9sin²θ) dθ = (1/9) ∫ csc²(θ) dθ
5. INTEGRATE WITH RESPECT TO θ:
(1/9) * (-cot(θ)) + C
6. CONVERT BACK TO x: Use a reference triangle.
From x = 3sin(θ), we have sin(θ) = x/3 (opposite/hypotenuse).
The adjacent side is √(3²-x²) = √(9-x²).
Therefore, cot(θ) = adjacent/opposite = √(9-x²)/x.
Final Answer: -(1/9) * [√(9-x²)/x] + C
</code>
</div>
</div>
<!-- MODULE 5: PARTIAL FRACTIONS -->
<div class="module">
<h2>Module 5: Partial Fraction Decomposition</h2>
<div class="theorem-box">
<strong>Theorem:</strong> Any rational function <span class="math">P(x)/Q(x)</span> where degree(P) < degree(Q) can be rewritten as a sum of simpler fractions whose denominators are the factors of <span class="math">Q(x)</span>.
</div>
<h3>Strategy:</h3>
<ul class="concept-list">
<li>If degree(P) >= degree(Q), perform polynomial long division first.</li>
<li>Factor the denominator <span class="math">Q(x)</span> completely.</li>
<li>For each linear factor <span class="math">(x-r)</span>, create a term <span class="math">A/(x-r)</span>.</li>
<li>For repeated linear factors <span class="math">(x-r)ⁿ</span>, create terms for each power: <span class="math">A₁/(x-r) + A₂/(x-r)² + ...</span></li>
<li>For irreducible quadratic factors <span class="math">(ax²+bx+c)</span>, create a term <span class="math">(Ax+B)/(ax²+bx+c)</span>.</li>
</ul>
<div class="example" onclick="toggleSolution(this)">Example 5.1: Evaluate <span class="integral">∫</span> <span class="frac"><span class="numer"><span class="math">x</span>+1</span><span class="denom"><span class="math">x</span>(<span class="math">x</span>-1)<sup>2</sup></span></span> d<span class="math">x</span></div>
<div class="solution">
<code>
1. SET UP THE DECOMPOSITION:
(x+1)/[x(x-1)²] = A/x + B/(x-1) + C/(x-1)²
2. CLEAR THE DENOMINATOR by multiplying through:
x+1 = A(x-1)² + Bx(x-1) + Cx
3. SOLVE FOR COEFFICIENTS (Heaviside Cover-up Method):
Let x=1: 1+1 = C(1) => C = 2
Let x=0: 0+1 = A(-1)² => A = 1
4. SOLVE FOR THE REMAINING COEFFICIENT (B):
Expand and equate the coefficients of the x² term.
x+1 = A(x²-2x+1) + B(x²-x) + Cx
x+1 = (A+B)x² + (-2A-B+C)x + A
The coefficient of x² on the left is 0.
0 = A+B => 0 = 1+B => B = -1
5. REWRITE THE INTEGRAL with the simpler fractions:
∫ [ 1/x - 1/(x-1) + 2/(x-1)² ] dx
6. INTEGRATE TERM BY TERM:
ln|x| - ln|x-1| + 2 * (-1/(x-1)) + C
ln|x| - ln|x-1| - 2/(x-1) + C
</code>
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