2.3 Partielle Integration
Aus Online Mathematik Brückenkurs 2
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To integrate products, one sometimes can make use of a method known as ''integration by parts''. The method is based on the reverse use of the rules for differentiation of products. If <math>u</math> and <math>v</math> are two differentiable functions then the rule for products gives | To integrate products, one sometimes can make use of a method known as ''integration by parts''. The method is based on the reverse use of the rules for differentiation of products. If <math>u</math> and <math>v</math> are two differentiable functions then the rule for products gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>D\,(\,u\, v) = u^{\,\prime} \, v + u \, v'\,\mbox{.}</math>}} |
Now if one integrates both sides one gets | Now if one integrates both sides one gets | ||
| - | {{ | + | {{Abgesetzte Formel||<math>u \, v = \int (\,u^{\,\prime} \, v + u \, v'\,)\,dx = \int u^{\,\prime} \, v\,dx + \int u\, v'\,dx</math>}} |
which, after re-ordering, becomes the formula for integration by parts. | which, after re-ordering, becomes the formula for integration by parts. | ||
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<div class="regel"> | <div class="regel"> | ||
'''Integration by parts:''' | '''Integration by parts:''' | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int u \, v'\,dx = u \, v - \int u^{\,\prime} \, v\,dx\,\mbox{.}</math>}} |
</div> | </div> | ||
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If one chooses <math>u=\sin x</math> and <math>v'=x</math> one gets <math>u'=\cos x</math> and <math>v=x^2/2</math>, and the formula for integration by parts gives | If one chooses <math>u=\sin x</math> and <math>v'=x</math> one gets <math>u'=\cos x</math> and <math>v=x^2/2</math>, and the formula for integration by parts gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int x \, \sin x \, dx = \frac{x^2}{2} \, \sin x - \int \frac{x^2}{2} \, \cos x \, dx\,\mbox{.}</math>}} |
The new integral on the right-hand side in this case is not easier than the original integral. | The new integral on the right-hand side in this case is not easier than the original integral. | ||
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If, instead, one chooses <math>u=x</math> and <math>v'=\sin x</math> then <math>u'=1</math> and <math>v=-\cos x</math>, and | If, instead, one chooses <math>u=x</math> and <math>v'=\sin x</math> then <math>u'=1</math> and <math>v=-\cos x</math>, and | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int x \, \sin x \, dx = - x \, \cos x - \int - 1 \times \cos x \, dx = - x\cos x + \sin x + C\,\mbox{.}</math>}} |
</div> | </div> | ||
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Put <math>u=\ln x</math> and <math>v'=x^2</math>, since differentiation eliminates the logarithm when we carry out an integration by parts: <math>u'=1/x</math> and <math>v=x^3/3</math>. This gives us | Put <math>u=\ln x</math> and <math>v'=x^2</math>, since differentiation eliminates the logarithm when we carry out an integration by parts: <math>u'=1/x</math> and <math>v=x^3/3</math>. This gives us | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align*}\int x^2 \, \ln x \, dx &= \frac {x^3}{3} \, \ln x - \int \frac{x^3}{3} \, \frac{1}{x} \, dx = \frac {x^3}{3} \, \ln x - \frac{1}{3} \int x^2 \, dx\\[4pt] &= \frac{x^3}{3} \, \ln x - \frac{1}{3} \, \frac{x^3}{3} + C = \tfrac{1}{3}x^3 ( \ln x - \tfrac{1}{3} ) + C\,\mbox{.}\end{align*}</math>}} |
</div> | </div> | ||
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Put <math>u=x^2</math> and <math>v'=e^x</math>, which gives <math>u'=2x</math> and <math>v=e^x</math>; integration by parts gives | Put <math>u=x^2</math> and <math>v'=e^x</math>, which gives <math>u'=2x</math> and <math>v=e^x</math>; integration by parts gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math> \int x^2 e^x \, dx = x^2 e^x - \int 2x\,e^x \, dx\,\mbox{.}</math>}} |
This requires further integration by parts to solve the new integral <math>\,\int 2x\,e^x \, dx</math>. We choose in this case <math>u=2x</math> and <math>v'=e^x</math>, which gives <math>u'=2</math> and <math>v=e^x</math>: | This requires further integration by parts to solve the new integral <math>\,\int 2x\,e^x \, dx</math>. We choose in this case <math>u=2x</math> and <math>v'=e^x</math>, which gives <math>u'=2</math> and <math>v=e^x</math>: | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int 2x\,e^x \, dx = 2x\,e^x - \int 2 e^x \, dx = 2x\,e^x - 2 e^x + C\,\mbox{.}</math>}} |
The original integral thus becomes | The original integral thus becomes | ||
| - | {{ | + | {{Abgesetzte Formel||<math> \int x^2 e^x \, dx = x^2 e^x - 2x\,e^x + 2 e^x + C\,\mbox{.}</math>}} |
</div> | </div> | ||
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In the first integration by parts, we have chosen to integrate the factor <math>e^x</math> and differentiate the factor <math>\cos x</math>, | In the first integration by parts, we have chosen to integrate the factor <math>e^x</math> and differentiate the factor <math>\cos x</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align*}\int e^x \cos x \, dx &= e^x \, \cos x - \int e^x \,(-\sin x) \, dx\\[10pt] &= e^x \cos x + \int e^x \sin x \, dx\,\mbox{.}\end{align*}</math>}} |
The result of this is that we essentially have replaced the factor <math>\cos x</math> by <math>\sin x</math> in the integral. If we therefore use integration by parts once again (integrate the <math>e^x</math> and differentiate the <math>\sin x</math>) we get | The result of this is that we essentially have replaced the factor <math>\cos x</math> by <math>\sin x</math> in the integral. If we therefore use integration by parts once again (integrate the <math>e^x</math> and differentiate the <math>\sin x</math>) we get | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx\,\mbox{.}</math>}} |
Thus the original integral appears here again. Summarising we have: | Thus the original integral appears here again. Summarising we have: | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int e^x \cos x \, dx = e^x \cos x + e^x \sin x - \int e^x \cos x \, dx</math>}} |
and collecting the integrals to one side gives | and collecting the integrals to one side gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int e^x \cos x \, dx = {\textstyle\frac{1}{2}}e^x ( \cos x + \sin x) + C\,\mbox{.}</math>}} |
Although integration by parts in this case did not lead to an easier integral, we arrived at an equation in which the original integral could be ”solved for”. This is not entirely unusual when the integrand is a product of trigonometric functions and / or exponential functions. | Although integration by parts in this case did not lead to an easier integral, we arrived at an equation in which the original integral could be ”solved for”. This is not entirely unusual when the integrand is a product of trigonometric functions and / or exponential functions. | ||
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The integral can be rewritten as | The integral can be rewritten as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int_{0}^{1} \frac{2x}{e^x} \, dx = \int_{0}^{1} 2x \, e^{-x} \, dx\,\mbox{.}</math>}} |
Substitute <math>u=2x</math> and <math>v'=e^{-x}</math>, and integrate by parts | Substitute <math>u=2x</math> and <math>v'=e^{-x}</math>, and integrate by parts | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align*}\int_{0}^{1} 2x \, e^{-x} \, dx &= \Bigl[\,-2x\,e^{-x}\,\Bigr]_{0}^{1} + \int_{0}^{1} 2 e^{-x}\,dx\\[4pt] &= \Bigl[\,-2x e^{-x}\,\Bigr]_{0}^{1} + \Bigl[\,-2 e^{-x}\, \Bigr]_{0}^{1}\\[4pt] &= (-2 \, e^{-1}) - 0 + (- 2\, e^{-1}) - (-2)\\[4pt] &= - \frac{2}{e} - \frac{2}{e} + 2 = 2 - \frac{4}{e}\,\mbox{.}\end{align*}</math>}} |
</div> | </div> | ||
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We start by performing a substitution <math>u=\sqrt{x}</math> which gives <math>du=dx/2\sqrt{x} = dx/2u</math>, that is, <math>dx = 2u\,du\,</math>, | We start by performing a substitution <math>u=\sqrt{x}</math> which gives <math>du=dx/2\sqrt{x} = dx/2u</math>, that is, <math>dx = 2u\,du\,</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int \ln \sqrt{x} \, dx = \int \ln u \times 2u \, du\,\mbox{.}</math>}} |
Then we integrate by parts. Put <math>f=\ln u</math> <math>g'=2u</math>, which gives | Then we integrate by parts. Put <math>f=\ln u</math> <math>g'=2u</math>, which gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align*}\int \ln u \times 2u \, du &= u^2 \ln u - \int u^2 \, \frac{1}{u} \, du = u^2 \ln u - \int u\, du\\[4pt] &= u^2 \ln u - \frac{u^2}{2} + C = x \ln \sqrt{x} - \frac {x}{2} + C\\[4pt] &= x \bigl( \ln \sqrt{x} - \tfrac{1}{2} \bigr) + C\,\mbox{.}\end{align*}</math>}} |
''Note.'' An alternative approach is to rewrite the initial integrand as <math>\ln\sqrt{x} = \tfrac{1}{2}\ln x</math> and then integrate by parts the product <math>\tfrac{1}{2}\,\ln x</math>. | ''Note.'' An alternative approach is to rewrite the initial integrand as <math>\ln\sqrt{x} = \tfrac{1}{2}\ln x</math> and then integrate by parts the product <math>\tfrac{1}{2}\,\ln x</math>. | ||
</div> | </div> | ||
Version vom 12:49, 10. Mär. 2009
| Theory | Exercises |
Contents:
- Integration by parts.
Learning outcomes:
After this section, you will have learned to:
- Understand the derivation of the formula for integration by parts.
- Solve problems about integration that require integration by parts, followed by a substitution (or vice versa).
Integration by parts
To integrate products, one sometimes can make use of a method known as integration by parts. The method is based on the reverse use of the rules for differentiation of products. If \displaystyle u and \displaystyle v are two differentiable functions then the rule for products gives
| \displaystyle D\,(\,u\, v) = u^{\,\prime} \, v + u \, v'\,\mbox{.} |
Now if one integrates both sides one gets
| \displaystyle u \, v = \int (\,u^{\,\prime} \, v + u \, v'\,)\,dx = \int u^{\,\prime} \, v\,dx + \int u\, v'\,dx |
which, after re-ordering, becomes the formula for integration by parts.
Integration by parts:
| \displaystyle \int u \, v'\,dx = u \, v - \int u^{\,\prime} \, v\,dx\,\mbox{.} |
This means in practice that one integrates a product of functions by calling one factor \displaystyle u and the other \displaystyle v', and then replaces the integral \displaystyle \,\int u \, v'\,dx\ , with the integral \displaystyle \,\int u^{\,\prime} \, v\,dx\,\mbox{,}\ , which one hopes will be easier. Here, \displaystyle v is any antiderivative (primitive function) of \displaystyle v' (by preference, the simplest) and \displaystyle u' is the derivative of \displaystyle u.
It is important to note that the method does not always lead to an integral that is easier than the original. It may also be crucial how one chooses the functions \displaystyle u and \displaystyle v', as the following example shows.
Example 1
Determine the integral \displaystyle \,\int x \, \sin x \, dx\,.
If one chooses \displaystyle u=\sin x and \displaystyle v'=x one gets \displaystyle u'=\cos x and \displaystyle v=x^2/2, and the formula for integration by parts gives
| \displaystyle \int x \, \sin x \, dx = \frac{x^2}{2} \, \sin x - \int \frac{x^2}{2} \, \cos x \, dx\,\mbox{.} |
The new integral on the right-hand side in this case is not easier than the original integral.
If, instead, one chooses \displaystyle u=x and \displaystyle v'=\sin x then \displaystyle u'=1 and \displaystyle v=-\cos x, and
| \displaystyle \int x \, \sin x \, dx = - x \, \cos x - \int - 1 \times \cos x \, dx = - x\cos x + \sin x + C\,\mbox{.} |
Example 2
Determine the integral \displaystyle \ \int x^2 \, \ln x \, dx\,.
Put \displaystyle u=\ln x and \displaystyle v'=x^2, since differentiation eliminates the logarithm when we carry out an integration by parts: \displaystyle u'=1/x and \displaystyle v=x^3/3. This gives us
| \displaystyle \begin{align*}\int x^2 \, \ln x \, dx &= \frac {x^3}{3} \, \ln x - \int \frac{x^3}{3} \, \frac{1}{x} \, dx = \frac {x^3}{3} \, \ln x - \frac{1}{3} \int x^2 \, dx\\[4pt] &= \frac{x^3}{3} \, \ln x - \frac{1}{3} \, \frac{x^3}{3} + C = \tfrac{1}{3}x^3 ( \ln x - \tfrac{1}{3} ) + C\,\mbox{.}\end{align*} |
Example 3
Determine the integral \displaystyle \ \int x^2 e^x \, dx\,.
Put \displaystyle u=x^2 and \displaystyle v'=e^x, which gives \displaystyle u'=2x and \displaystyle v=e^x; integration by parts gives
| \displaystyle \int x^2 e^x \, dx = x^2 e^x - \int 2x\,e^x \, dx\,\mbox{.} |
This requires further integration by parts to solve the new integral \displaystyle \,\int 2x\,e^x \, dx. We choose in this case \displaystyle u=2x and \displaystyle v'=e^x, which gives \displaystyle u'=2 and \displaystyle v=e^x:
| \displaystyle \int 2x\,e^x \, dx = 2x\,e^x - \int 2 e^x \, dx = 2x\,e^x - 2 e^x + C\,\mbox{.} |
The original integral thus becomes
| \displaystyle \int x^2 e^x \, dx = x^2 e^x - 2x\,e^x + 2 e^x + C\,\mbox{.} |
Example 4
Determine the integral \displaystyle \ \int e^x \cos x \, dx\,.
In the first integration by parts, we have chosen to integrate the factor \displaystyle e^x and differentiate the factor \displaystyle \cos x,
| \displaystyle \begin{align*}\int e^x \cos x \, dx &= e^x \, \cos x - \int e^x \,(-\sin x) \, dx\\[10pt] &= e^x \cos x + \int e^x \sin x \, dx\,\mbox{.}\end{align*} |
The result of this is that we essentially have replaced the factor \displaystyle \cos x by \displaystyle \sin x in the integral. If we therefore use integration by parts once again (integrate the \displaystyle e^x and differentiate the \displaystyle \sin x) we get
| \displaystyle \int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx\,\mbox{.} |
Thus the original integral appears here again. Summarising we have:
| \displaystyle \int e^x \cos x \, dx = e^x \cos x + e^x \sin x - \int e^x \cos x \, dx |
and collecting the integrals to one side gives
| \displaystyle \int e^x \cos x \, dx = {\textstyle\frac{1}{2}}e^x ( \cos x + \sin x) + C\,\mbox{.} |
Although integration by parts in this case did not lead to an easier integral, we arrived at an equation in which the original integral could be ”solved for”. This is not entirely unusual when the integrand is a product of trigonometric functions and / or exponential functions.
Example 5
Determine the integral \displaystyle \ \int_{0}^{1} \frac{2x}{e^x} \, dx\,.
The integral can be rewritten as
| \displaystyle \int_{0}^{1} \frac{2x}{e^x} \, dx = \int_{0}^{1} 2x \, e^{-x} \, dx\,\mbox{.} |
Substitute \displaystyle u=2x and \displaystyle v'=e^{-x}, and integrate by parts
| \displaystyle \begin{align*}\int_{0}^{1} 2x \, e^{-x} \, dx &= \Bigl[\,-2x\,e^{-x}\,\Bigr]_{0}^{1} + \int_{0}^{1} 2 e^{-x}\,dx\\[4pt] &= \Bigl[\,-2x e^{-x}\,\Bigr]_{0}^{1} + \Bigl[\,-2 e^{-x}\, \Bigr]_{0}^{1}\\[4pt] &= (-2 \, e^{-1}) - 0 + (- 2\, e^{-1}) - (-2)\\[4pt] &= - \frac{2}{e} - \frac{2}{e} + 2 = 2 - \frac{4}{e}\,\mbox{.}\end{align*} |
Example 6
Determine the integral \displaystyle \ \int \ln \sqrt{x} \ dx\,.
We start by performing a substitution \displaystyle u=\sqrt{x} which gives \displaystyle du=dx/2\sqrt{x} = dx/2u, that is, \displaystyle dx = 2u\,du\,,
| \displaystyle \int \ln \sqrt{x} \, dx = \int \ln u \times 2u \, du\,\mbox{.} |
Then we integrate by parts. Put \displaystyle f=\ln u \displaystyle g'=2u, which gives
| \displaystyle \begin{align*}\int \ln u \times 2u \, du &= u^2 \ln u - \int u^2 \, \frac{1}{u} \, du = u^2 \ln u - \int u\, du\\[4pt] &= u^2 \ln u - \frac{u^2}{2} + C = x \ln \sqrt{x} - \frac {x}{2} + C\\[4pt] &= x \bigl( \ln \sqrt{x} - \tfrac{1}{2} \bigr) + C\,\mbox{.}\end{align*} |
Note. An alternative approach is to rewrite the initial integrand as \displaystyle \ln\sqrt{x} = \tfrac{1}{2}\ln x and then integrate by parts the product \displaystyle \tfrac{1}{2}\,\ln x.
