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Calculus - Indefinite Integrals: Integration by Substitution, by Parts, and by Partial Fractions

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Indefinite Integration as Anti-differentiation: Integration is the reverse process of differentiation. Visually, the indefinite integral of a function represents a 'family of curves.' Each curve in the family, defined by y=F(x)+Cy = F(x) + C, is a vertical translation of the others. This means that at any specific value of xx, the tangent lines to all these curves are parallel, having the same slope f(x)f(x).

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Integration by Substitution (Change of Variable): This method is used when the integrand contains a function and its derivative, effectively reversing the Chain Rule. By substituting u=g(x)u = g(x), we transform the integral ∫f(g(x))g′(x)dx\int f(g(x))g'(x) dx into a simpler form ∫f(u)du\int f(u) du. Visually, this can be thought of as stretching or compressing the x-axis to simplify the area under the curve.

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Integration by Parts: Based on the product rule of differentiation, this technique is used for the product of two functions. It is defined as ∫udv=uv−∫vdu\int u dv = uv - \int v du. To choose which function is uu and which is dvdv, we follow the ILATE rule: Inverse Trigonometric, Logarithmic, Algebraic, Trigonometric, and Exponential. This method effectively breaks a complex product into a manageable boundary term and a simpler integral.

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Integration using Partial Fractions: This algebraic technique decomposes a complex rational function P(x)Q(x)\frac{P(x)}{Q(x)} (where the degree of P(x)P(x) is less than Q(x)Q(x)) into a sum of simpler fractions. For example, a denominator with distinct linear factors like (x−a)(x−b)(x-a)(x-b) is split into terms like Ax−a+Bx−b\frac{A}{x-a} + \frac{B}{x-b}. Visually, it decomposes a single complex rate of change into its constituent simpler components.

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The Constant of Integration (CC): Because the derivative of any constant is zero, an indefinite integral always includes an arbitrary constant CC. This represents the vertical position of the function on a Cartesian plane. Without specific boundary conditions (initial values), the exact vertical level of the curve remains 'indefinite'.

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Integrals of Special Rational Functions: Specific forms such as ∫dxx2−a2\int \frac{dx}{x^2 - a^2} or ∫dxa2−x2\int \frac{dx}{\sqrt{a^2 - x^2}} have standardized logarithmic or inverse trigonometric solutions. Graphically, the square root forms often relate to the geometry of circles and hyperbolas.

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Linearity Property: Integration is a linear operator, meaning ∫[af(x)+bg(x)]dx=a∫f(x)dx+b∫g(x)dx\int [af(x) + bg(x)] dx = a\int f(x) dx + b\int g(x) dx. This allows us to integrate complex polynomials term-by-term, visualizing the total area as the sum of smaller, simpler geometric areas.

📐Formulae

∫xndx=xn+1n+1+C,(n≠−1)\int x^n dx = \frac{x^{n+1}}{n+1} + C, (n \neq -1)

∫1xdx=ln⁡∣x∣+C\int \frac{1}{x} dx = \ln|x| + C

∫exdx=ex+C\int e^x dx = e^x + C

∫udvdxdx=uv−∫vdudxdx\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx

∫dxx2+a2=1atan⁡−1(xa)+C\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C

∫dxa2−x2=sin⁡−1(xa)+C\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + C

∫tan⁡xdx=ln⁡∣sec⁡x∣+C\int \tan x dx = \ln|\sec x| + C

∫dxx2−a2=12aln⁡∣x−ax+a∣+C\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln|\frac{x-a}{x+a}| + C

💡Examples

Problem 1:

Evaluate ∫xcos⁡(x2)dx\int x \cos(x^2) dx using the substitution method.

Solution:

  1. Let u=x2u = x^2.
  2. Differentiate both sides: dudx=2x\frac{du}{dx} = 2x, which gives du=2xdxdu = 2x dx or xdx=12dux dx = \frac{1}{2} du.
  3. Substitute these into the integral: ∫cos⁡(u)⋅12du\int \cos(u) \cdot \frac{1}{2} du.
  4. Factor out the constant: 12∫cos⁡(u)du\frac{1}{2} \int \cos(u) du.
  5. Integrate: 12sin⁡(u)+C\frac{1}{2} \sin(u) + C.
  6. Substitute back u=x2u = x^2: 12sin⁡(x2)+C\frac{1}{2} \sin(x^2) + C.

Explanation:

This problem uses substitution because the derivative of the inner function x2x^2 (which is 2x2x) is present as a factor in the integrand. By changing the variable to uu, we convert a composite function into a basic trigonometric integral.

Problem 2:

Evaluate ∫xexdx\int x e^x dx using Integration by Parts.

Solution:

  1. Using ILATE, choose u=xu = x (Algebraic) and dv=exdxdv = e^x dx (Exponential).
  2. Find dudu: du=dxdu = dx.
  3. Find vv: v=∫exdx=exv = \int e^x dx = e^x.
  4. Apply the formula ∫udv=uv−∫vdu\int u dv = uv - \int v du: ∫xexdx=(x)(ex)−∫exdx\int x e^x dx = (x)(e^x) - \int e^x dx.
  5. Evaluate the remaining integral: xex−ex+Cxe^x - e^x + C.
  6. Factor out exe^x: ex(x−1)+Ce^x(x - 1) + C.

Explanation:

Integration by parts is chosen here because we have a product of two different types of functions (algebraic and exponential). The method reduces the power of xx until the integral becomes a simple exponential form.