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Calculus - Further integration by substitution (HL)

Grade 12IB_AA

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

🔑Concepts

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Integration by substitution is the inverse process of the chain rule for differentiation, used to simplify an integrand by changing the variable of integration.

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The substitution u=g(x)u = g(x) is typically chosen such that its derivative g′(x)g'(x) is also present in the integrand, or to simplify a nested function.

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When performing a substitution in a definite integral, the limits of integration MUST be changed from xx-values to uu-values using the substitution formula u=g(x)u = g(x).

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For HL, substitutions may involve more complex algebraic manipulations, such as solving for xx in terms of uu to replace remaining xx terms in the integrand (e.g., if u=x+1u = x+1, then x=u−1x = u-1).

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Trigonometric substitutions are a subset of this method, where terms like a2−x2\sqrt{a^2 - x^2} suggest x=asin⁡(θ)x = a \sin(\theta) and terms like a2+x2a^2 + x^2 suggest x=atan⁡(θ)x = a \tan(\theta).

📐Formulae

∫f(g(x))g′(x)dx=∫f(u)du, where u=g(x)\int f(g(x)) g'(x) dx = \int f(u) du, \text{ where } u = g(x)

∫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

dudx=g′(x)  ⟹  dx=dug′(x)\frac{du}{dx} = g'(x) \implies dx = \frac{du}{g'(x)}

∫f′(x)f(x)dx=ln⁡∣f(x)∣+C\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C

∫f(ax+b)dx=1aF(ax+b)+C, where F′(x)=f(x)\int f(ax+b) dx = \frac{1}{a} F(ax+b) + C, \text{ where } F'(x) = f(x)

💡Examples

Problem 1:

Evaluate the indefinite integral ∫xx−1dx\int x \sqrt{x-1} dx using the substitution u=x−1u = x-1.

Solution:

Let u=x−1u = x - 1. Then dudx=1\frac{du}{dx} = 1, which means dx=dudx = du. Also, if u=x−1u = x - 1, then x=u+1x = u + 1. Substitute these into the integral: ∫(u+1)udu=∫(u+1)u1/2du\int (u+1) \sqrt{u} du = \int (u+1) u^{1/2} du ∫(u3/2+u1/2)du=u5/25/2+u3/23/2+C\int (u^{3/2} + u^{1/2}) du = \frac{u^{5/2}}{5/2} + \frac{u^{3/2}}{3/2} + C 25u5/2+23u3/2+C\frac{2}{5}u^{5/2} + \frac{2}{3}u^{3/2} + C Substitute back u=x−1u = x - 1: 25(x−1)5/2+23(x−1)3/2+C\frac{2}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2} + C

Explanation:

In this case, the derivative of the inner function was 1, but we still needed to substitute the xx term outside the square root by rearranging the substitution equation.

Problem 2:

Evaluate the definite integral ∫01ex1+exdx\int_{0}^{1} \frac{e^x}{1 + e^x} dx.

Solution:

Let u=1+exu = 1 + e^x. Then dudx=ex\frac{du}{dx} = e^x, so du=exdxdu = e^x dx. Change the limits: When x=0,u=1+e0=1+1=2x = 0, u = 1 + e^0 = 1 + 1 = 2. When x=1,u=1+e1=1+ex = 1, u = 1 + e^1 = 1 + e. The integral becomes: ∫21+e1udu\int_{2}^{1+e} \frac{1}{u} du [ln⁡∣u∣]21+e=ln⁡(1+e)−ln⁡(2)[ \ln|u| ]_{2}^{1+e} = \ln(1+e) - \ln(2) Using log laws: ln⁡(1+e2)\ln\left(\frac{1+e}{2}\right)

Explanation:

This example uses the form ∫f′(x)f(x)dx\int \frac{f'(x)}{f(x)} dx. Notice how the limits were updated to the uu domain, so we do not need to substitute back to xx at the end.

Problem 3:

Find ∫sin⁡3(x)cos⁡(x)dx\int \sin^3(x) \cos(x) dx.

Solution:

Let u=sin⁡(x)u = \sin(x). Then dudx=cos⁡(x)\frac{du}{dx} = \cos(x), so du=cos⁡(x)dxdu = \cos(x) dx. Substitute into the integral: ∫u3du=u44+C\int u^3 du = \frac{u^4}{4} + C Substitute back u=sin⁡(x)u = \sin(x): sin⁡4(x)4+C\frac{\sin^4(x)}{4} + C

Explanation:

For integrals involving powers of trigonometric functions, look for the function whose derivative is also present. Here, the derivative of sin⁡(x)\sin(x) is cos⁡(x)\cos(x).