Integration by Parts

It is not possible to integrate all functions algebraically. But, there are some techniques which enable us to integrate more complex functions.

Integration by parts - partielle integration

This method uses the product rule (Produktregel).

$${\frac{d}{dx}(uv) = v \frac{du}{dx} + u \frac{dv}{dx}}$$

Integrating both sides we get.

$${\Rightarrow uv = \int{v \frac{du}{dx},\mathrm{d}x} + \int{u \frac{dv}{dx},\mathrm{d}x}}$$

$${ \int{u \frac{dv}{dx},\mathrm{d}x} = uv - \int{v \frac{du}{dx},\mathrm{d}x}}$$

Using the notation as in Cornelsen book

$${(uv)' = u' \cdot v + u \cdot v'}$$

$${\Rightarrow uv = \int{u' \cdot v ,\mathrm{d}x} + \int{u \cdot v' ,\mathrm{d}x}}$$

Nach umformung:

$${\int{u' \cdot v ,\mathrm{d}x} = uv - \int{u \cdot v' ,\mathrm{d}x}}$$

example

Berechnen Sie das unbestimmte Integral ${\int{\mathrm{e}^x \cdot x\,\mathrm{d}x}}$.

Solution: Let ${u'(x) = \mathrm{e}^x}$ and ${v(x) = x}$. Then ${u(x) = \mathrm{e}^x}$ and ${v'(x) = 1}$

So,

$${\begin{align}\Rightarrow \int{\mathrm{e}^x \cdot x,\mathrm{d}x} & = \mathrm{e}^x \cdot x - \int{\mathrm{e}^x \cdot 1 ,\mathrm{d}x}\newline & = \mathrm{e}^x \cdot x - \mathrm{e}^x + \mathrm{C} \newline & = \mathrm{e}^x (x - 1) + \mathrm{C} \end{align}}$$

task

Find the integrals.

no 1

$${\int{x \cdot \mathrm{e}^{-x},\mathrm{d}x}}$$ $${\int{x \cdot \mathrm{e}^{3x},\mathrm{d}x}}$$ $${\int{x \cdot \mathrm{e}^{ax},\mathrm{d}x}}$$ solution videos

no 2

$${\int{x \cos{x},\mathrm{d}x}}$$ $${\int{x \cos{3x},\mathrm{d}x}}$$ $${\int{x \cos{ax},\mathrm{d}x}}$$

no 3 - use integration by parts twice to evaluate these.

$${\text{(a)}~~\int{x^2 \mathrm{e}^{x}\,\mathrm{d}x}}$$ $${\text{(b)}~~\int{x^2 \sin{x}\,\mathrm{d}x}}$$

no 4

Work out each of these difinite integrals. Sketch diagrams to show the areas you have found and check that your answers seem reasonable. $${\text{(a)}~~\int\limits_{-1}^{0}{x \sin{2x},\mathrm{d}x}}$$

$${\text{(b)}~~\int\limits_{-3}^{0}{2x,\mathrm{e}^{0.5x},\mathrm{d}x}}$$

no 5

(a)

Find ${\int{x^2\, \ln{x}\,\mathrm{d}x}}$ using integegration by parts with ${u = \ln{x}}$, ${v^\prime = x^2}$

(b)

Show that $${\left[ x\ln{x} - x\right]' = \ln{x}}$$

(c)

Find ${\int{x^2\, \ln{x}\,\mathrm{d}x}}$ using integegration by parts with ${u = x^2 }$, ${v^\prime = \ln{x}}$

When the initial integral “reappears” in the process

example

Integrate ${\int{\cos x \cdot \sin x\, \mathrm{d}x}}$ (Berechnen Sie das Integral)

solution

Now, integrate ${\int{\cos^2 x \, \mathrm{d}x}}$. Remember that ${\cos^2 x + \sin^2 x = 1}$ (Pythagoras' theorem)

exercises on trigonometry