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Calculus - Techniques of Integration (Substitution, By Parts)

Grade 12IB

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

🔑Concepts

Integration by Substitution (uu-substitution) is the reverse process of the Chain Rule. It is used when an integrand contains a function g(x)g(x) and its derivative g(x)g'(x). Visually, this technique can be viewed as a coordinate transformation that simplifies a complex area calculation on the xx-axis into a simpler one on a uu-axis, effectively 'straightening' the function's internal composition.

When applying substitution to definite integrals, it is essential to transform the limits of integration. If the original limits are x=ax = a and x=bx = b, the new limits become u=g(a)u = g(a) and u=g(b)u = g(b). This ensures that the area under the curve is calculated over the correct interval in the transformed coordinate system without needing to substitute back to xx.

Integration by Parts (IBP) is the integration equivalent of the Product Rule for differentiation. Geometrically, if you consider a curve in a uu-vv plane, the formula relates the area under the curve u(v)u(v) to the area of a rectangle uvuv minus the area relative to the other axis. It is most effective when the integrand is a product of two dissimilar functions, such as an algebraic and a trigonometric function.

The LIATE rule is a visual mnemonic used to choose which part of the integrand should be assigned to uu in Integration by Parts. The priority order is: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. By choosing uu following this hierarchy, the resulting integral vdu\int v du is generally simpler than the original, as the functions higher in the list simplify significantly when differentiated.

The 'Logarithmic Pattern' recognizes integrals in the form f(x)f(x)dx\int \frac{f'(x)}{f(x)} dx. On a graph, this represents the area under a reciprocal-style function where the numerator is the exact rate of change of the denominator. The result is always lnf(x)+C\ln|f(x)| + C.

Repeated Integration by Parts is required when the algebraic part of the integrand has a degree higher than 1 (e.g., x2sin(x)x^2 \sin(x)). In some cases, such as excos(x)dx\int e^x \cos(x) dx, the process 'loops' and returns to the original integral form, allowing you to solve for the integral algebraically as if it were a variable in an equation.

Visualizing the 'Differential' dudu: In substitution, the term du=g(x)dxdu = g'(x) dx represents the change in the new variable relative to xx. Visually, if g(x)g(x) is steep, a small change in xx results in a large change in uu, and the substitution accounts for this stretching or shrinking of the horizontal units across the area being calculated.

📐Formulae

f(g(x))g(x)dx=f(u)du\int f(g(x))g'(x) dx = \int f(u) du

abf(g(x))g(x)dx=g(a)g(b)f(u)du\int_{a}^{b} f(g(x))g'(x) dx = \int_{g(a)}^{g(b)} f(u) du

udvdxdx=uvvdudxdx\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx

udv=uvvdu\int u dv = uv - \int v du

f(x)f(x)dx=lnf(x)+C\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C

tan(x)dx=lnsec(x)+C\int \tan(x) dx = \ln|\sec(x)| + C

💡Examples

Problem 1:

Evaluate the indefinite integral x2ex3dx\int x^2 e^{x^3} dx.

Solution:

  1. Identify the internal function g(x)=x3g(x) = x^3. Its derivative is 3x23x^2, which matches the x2x^2 term outside except for a constant.
  2. Let u=x3u = x^3, then dudx=3x2\frac{du}{dx} = 3x^2, which gives dx=du3x2dx = \frac{du}{3x^2} or x2dx=13dux^2 dx = \frac{1}{3} du.
  3. Substitute these into the integral: eu13du\int e^u \cdot \frac{1}{3} du.
  4. Factor out the constant: 13eudu\frac{1}{3} \int e^u du.
  5. Integrate: 13eu+C\frac{1}{3} e^u + C.
  6. Substitute back for xx: 13ex3+C\frac{1}{3} e^{x^3} + C.

Explanation:

This problem uses uu-substitution. By recognizing that x2x^2 is a scalar multiple of the derivative of the exponent x3x^3, we can collapse the composite function into a basic exponential form.

Problem 2:

Find the integral xcos(x)dx\int x \cos(x) dx.

Solution:

  1. Use Integration by Parts. Following LIATE, let u=xu = x (Algebraic) and dv=cos(x)dxdv = \cos(x) dx (Trigonometric).
  2. Differentiate uu: du=dxdu = dx.
  3. Integrate dvdv: v=cos(x)dx=sin(x)v = \int \cos(x) dx = \sin(x).
  4. Apply the formula udv=uvvdu\int u dv = uv - \int v du: xcos(x)dx=xsin(x)sin(x)dx\int x \cos(x) dx = x \sin(x) - \int \sin(x) dx.
  5. Evaluate the remaining integral: sin(x)dx=cos(x)\int \sin(x) dx = -\cos(x).
  6. Combine results: xsin(x)(cos(x))+C=xsin(x)+cos(x)+Cx \sin(x) - (-\cos(x)) + C = x \sin(x) + \cos(x) + C.

Explanation:

This problem requires Integration by Parts because the integrand is a product of two different types of functions. Differentiating xx reduces it to a constant, making the second integral easy to solve.