integration by substitution

the chain rule of differentiation

Remember,

$${[f(g(x))]’ = f’(g(x)) \cdot g’(x)}$$

or in Leibniz notation,

$${\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot \frac{\mathrm{d}u}{\mathrm{d}x}}$$

which can also be written as

$${\frac{\mathrm{d}y}{\mathrm{d}u}=\frac{\mathrm{d}y}{\mathrm{d}x}\cdot \frac{\mathrm{d}x}{\mathrm{d}u}}$$

task

If $${\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x}{\sqrt{x^2+1}}}$$ and ${u = x^2 + 1}$, write down an expression for ${\frac{\mathrm{d}y}{\mathrm{d}x}}$ and hence find ${\frac{\mathrm{d}y}{\mathrm{d}u}}$ (a) in terms of ${x}$, (b) in terms of ${u}$.

solution

example

If $${\frac{\mathrm{d}y}{\mathrm{d}x}=x^2(2x+3)^4}$$ and ${u = 2x + 3}$, find terms ${\frac{\mathrm{d}y}{\mathrm{d}u}}$ in terms of ${u}$.

solution

example 1

Gesucht ist das unbestimmte Integral $${\int{(4x + 1)^3,\mathrm{d}x}}$$

1. Substitution

$${4x + 1 = z}$$

2. Differentiale

$${z^\prime = \frac{\mathrm{d}z}{\mathrm{d}x} =4}$$ $${\Rightarrow \mathrm{d}x = \frac{\mathrm{d}z}{4}}$$

3. Einsetzen von 1. und 2. ins Integral

$${\begin{align} \int{(4x + 1)^3,\mathrm{d}x} & = \int{z^3,\frac{\mathrm{d}z}{4}} \newline & = \frac{1}{16} z^4 + \mathrm{C}\end{align}}$$

4. Resubstitution

$${z = 4x + 1}$$ $${\begin{align} \int{(4x + 1)^3,\mathrm{d}x} = \frac{1}{16} (4x + 1)^4 + \mathrm{C}\end{align}}$$

example 2

Find $${y = \int{x \cos{(3x^2 + 5)},\mathrm{d}x}}$$

solution

$${z = 3x^2 + 5}$$ 2. $${z^\prime = \frac{\mathrm{d}z}{\mathrm{d}x} = 6x}$$ $${\Rightarrow \mathrm{d}x = \frac{\mathrm{d}z}{6x}}$$ 3. $${\begin{align} \int{x \cos{(3x^2 + 5)},\mathrm{d}x} & = \int{x \cos{(z)},\frac{\mathrm{d}z}{6x}} \newline & = \int{\cos{(z)},\frac{\mathrm{d}z}{6}} \newline & = \frac{1}{6} \sin{(z)} + \mathrm{C} \newline & = \frac{1}{6} \sin{(3x^2 + 5)} + \mathrm{C} \end{align}}$$

definite integrals

What happens to the upper and lower limit of the integrals?

If you substitute back after integrating, you keep the initial limits of the integral. And, if you decide not to substitute back after integrating, you have to adjust the limits by substituting them both into the epression for z (or u or whatever the new variable is called)

example 3 - difinite integral

Evaluate $${\int \limits_{0}^{2}{\frac{4x}{\sqrt{1 + 2x^2}}},\mathrm{d}x}$$

solution

method

First select the “inner function z” and the “outer function” so that

  1. substitution $${z = 1 + 2x^2}$$

  2. derivative (differentiale) $${z’ = \frac{\mathrm{d}z}{\mathrm{d}x} = 4x}$$ $${\Rightarrow \mathrm{d}x = \frac{\mathrm{d}z}{4x}}$$ $${\Rightarrow \mathrm{d}z = 4x \mathrm{d}z}$$

  3. converting the limits

    • lower limit: ${z = 1 + 2\cdot 0^2 = 1}$

    • upper limit: ${z = 1 + 2\cdot 2^2 = 9}$

  4. integration ${\int \limits_{0}^{2}{\frac{4x}{\sqrt{1 + 2x^2}}}\,\mathrm{d}x = \int \limits_{0}^{2}{\frac{1}{\sqrt{1 + 2x^2}} \cdot 4x}\,\mathrm{d}x = \int \limits_{1}^{9}{\frac{1}{\sqrt{z}}}\,\mathrm{d}z = \left[2 \sqrt {z} \right]_{1}^{9} = 4}$

Note You can resubstitute ${z = 1 + 2x^2}$ which would lead to the same result.

$${\int \limits_{0}^{2}{\frac{4x}{\sqrt{1 + 2x^2}}},\mathrm{d}x = … =\left[2 \sqrt {z} \right]_{1}^{9} = \left[2 \sqrt {1 + 2x^2} \right]_{0}^{2} = 4}$$

example 4 thumbnail

Find the integral of $${\int {\frac{x}{\sqrt{1 - x}}},\mathrm{d}x}$$ by substituting $z=\sqrt{1-x}$

Solution - click on image to enlarge.

example 5 - Using trigonometry and Pythagoras

Evaluate

$${\int \limits_{0}^{\frac{1}{2}}{\frac{1}{\sqrt{1 - x^2}}},\mathrm{d}x}$$

remember that $${\sin^2{x} + \cos^2{x} = 1}$$ $${\Rightarrow \cos{x} = \sqrt{1 - \sin^2{x}}}$$

So, we

  1. substitute ${x = \sin{t}}$

PDF with exercises

Get following PDF here: math_task_int_2

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