6ASubstitution

Integration by Substitution

When antidifferentiating, there is no chain rule, or product rule, or quotient rule. When the expression to be integrated is complicated, other methods must be used.

Use substitution when you see an expression within the integral that has had some other process done to it AND you recognize that the derivative of that expression also appears in the integral.

Example 1

math \displaystyle{\int} \dfrac {2x}{\sqrt {x^2+1}} dx math

Notice that //**x 2 + 1**// has had another process done to it and that the derivative of //**x 2 + 1** // is //**2x**// which also appears in the integral.

math \\ \text {Let } \; u = x^2 + 1 \\ \\ \text {so } \; \dfrac {du}{dx} = 2x \\ \\ \text {so } \; dx = \dfrac {du}{2x} math

Hence substitute **//u = x 2 + 1//** and substitute **//dx = du/2x//** math \displaystyle{\int} \dfrac {2x}{\sqrt {x^2+1}} dx = \displaystyle{\int} \dfrac {2x}{\sqrt {u}} \dfrac {du}{2x} math math \\ . \qquad \qquad \quad = \displaystyle{\int} \dfrac {1}{\sqrt {u}} du \\ \\ . \qquad \qquad \quad = \displaystyle{\int} u^{-\frac{1}{2}} du math math \\ . \qquad \qquad \quad = \dfrac {1}{\frac{1}{2}} u^{\frac{1}{2}} + c \\ \\ . \qquad \qquad \quad = 2 \sqrt{u} + c math Now substitute **//u = x 2 + 1//** back into the answer math \displaystyle{\int} \dfrac {2x}{\sqrt {x^2+1}} dx = 2 \sqrt{x^2 + 1} + c, \; \; x \in R math

Example 2

math \int { (x^2 - 1) \cos (3x - x^3) }dx math

math \\ \text{Let } \; u = 3x - x^3 \\ \\ \dfrac{du}{dx} = 3-3x^2 = -3(x^2 - 1) \\ \\ dx = \dfrac {du}{-3(x^2 - 1)} math

substitute **u** and **dx** into the integral math \int { (x^2 - 1) \cos (3x - x^3) }dx = \int { (x^2 - 1) \cos u \dfrac {du}{-3(x^2-1)} } math math \\ . \qquad \qquad \qquad \qquad \qquad \quad = -\dfrac {1}{3} \int {cos u} du \\ . \qquad \qquad \qquad \qquad \qquad \quad = -\dfrac {1}{3} \sin u + c math

Now substitute **u** back into the integral math \int { (x^2 - 1) \cos (3x - x^3) }dx = -\dfrac {1}{3} \sin (3x - x^3) + c math

**See also:** www.mathsonline.com.au Y12Extension --> Calculus --> Methods of Integration --> Lesson 1

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