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>f</math> and <math>g</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>f</math> and <math>g</math> are two differentiable functions then the rule for products gives | ||
| - | {{ | + | {{Displayed math||<math>D\,(\,f\cdot g) = f^{\,\prime} \cdot g + f \cdot g'\,\mbox{.}</math>}} |
Now if one integrates both sides one gets | Now if one integrates both sides one gets | ||
| - | {{ | + | {{Displayed math||<math>f \cdot g = \int (\,f^{\,\prime} \cdot g + f \cdot g'\,)\,dx = \int f^{\,\prime} \cdot g\,dx + \int f\cdot g'\,dx</math>}} |
or after re-ordering | or after re-ordering | ||
| - | {{ | + | {{Displayed math||<math>\int f^{\,\prime} \cdot g\,dx = f \cdot g - \int f \cdot g'\,dx\,\mbox{.}</math>}} |
This gives us the formula for integration by parts. | This gives us the formula for integration by parts. | ||
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<div class="regel"> | <div class="regel"> | ||
'''Integration by parts:''' | '''Integration by parts:''' | ||
| - | {{ | + | {{Displayed math||<math>\int f(x)\cdot g(x)\,dx = F(x) \cdot g(x) - \int F(x) \cdot g'(x)\,dx\,\mbox{.}</math>}} |
</div> | </div> | ||
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If one chooses <math>f=x</math> and <math>g=\sin x</math> one gets <math>F=x^2/2</math> and <math>g'=\cos x</math>, and the formula for integration by parts gives | If one chooses <math>f=x</math> and <math>g=\sin x</math> one gets <math>F=x^2/2</math> and <math>g'=\cos x</math>, and the formula for integration by parts gives | ||
| - | {{ | + | {{Displayed math||<math>\int x \cdot \sin x \, dx = \frac{x^2}{2} \cdot \sin x - \int \frac{x^2}{2} \cdot \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>f=\sin x</math> and <math>g=x</math> then <math>F=-\cos x</math> and <math>g'=1</math>, and | If, instead, one chooses <math>f=\sin x</math> and <math>g=x</math> then <math>F=-\cos x</math> and <math>g'=1</math>, and | ||
| - | {{ | + | {{Displayed math||<math>\int x \cdot \sin x \, dx = - x \cdot \cos x - \int - 1 \cdot \cos x \, dx = - x\cos x + \sin x + C\,\mbox{.}</math>}} |
</div> | </div> | ||
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Put <math>f=x^2</math> and <math>g=\ln x</math> since differentiation eliminates the logarithm when we carry out an integration by parts: <math>F=x^3/3</math> and <math>g'=1/x</math>. This gives us that | Put <math>f=x^2</math> and <math>g=\ln x</math> since differentiation eliminates the logarithm when we carry out an integration by parts: <math>F=x^3/3</math> and <math>g'=1/x</math>. This gives us that | ||
| - | {{ | + | {{Displayed math||<math>\begin{align*}\int x^2 \cdot \ln x \, dx &= \frac {x^3}{3} \cdot \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac {x^3}{3} \cdot \ln x - \frac{1}{3} \int x^2 \, dx\\[4pt] &= \frac{x^3}{3} \cdot \ln x - \frac{1}{3} \cdot \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>f=e^x</math> and <math>g=x^2</math>, which gives that <math>F=e^x</math> and <math>g'=2x</math>, and an integration by parts gives | Put <math>f=e^x</math> and <math>g=x^2</math>, which gives that <math>F=e^x</math> and <math>g'=2x</math>, and an integration by parts gives | ||
| - | {{ | + | {{Displayed math||<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>f=e^x</math> and <math>g=2x</math>, which gives <math>F=e^x</math> and <math>g'=2</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>f=e^x</math> and <math>g=2x</math>, which gives <math>F=e^x</math> and <math>g'=2</math> | ||
| - | {{ | + | {{Displayed math||<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 | ||
| - | {{ | + | {{Displayed math||<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 a first integration by parts, we have chosen to integrate the factor <math>e^x</math> and differentiate the factor <math>\cos x</math>, | In a first integration by parts, we have chosen to integrate the factor <math>e^x</math> and differentiate the factor <math>\cos x</math>, | ||
| - | {{ | + | {{Displayed math||<math>\begin{align*}\int e^x \cos x \, dx &= e^x \cdot \cos x - \int e^x \cdot(-\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 | ||
| - | {{ | + | {{Displayed math||<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: | ||
| - | {{ | + | {{Displayed math||<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 | ||
| - | {{ | + | {{Displayed math||<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 | ||
| - | {{ | + | {{Displayed math||<math>\int_{0}^{1} \frac{2x}{e^x} \, dx = \int_{0}^{1} 2x \cdot e^{-x} \, dx\,\mbox{.}</math>}} |
Substitute <math>f=e^{-x}</math> and <math>g=2x</math>, and integrate by parts | Substitute <math>f=e^{-x}</math> and <math>g=2x</math>, and integrate by parts | ||
| - | {{ | + | {{Displayed math||<math>\begin{align*}\int_{0}^{1} 2x \cdot 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 \cdot e^{-1}) - 0 + (- 2\cdot 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 variable 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 variable substitution <math>u=\sqrt{x}</math> which gives <math>du=dx/2\sqrt{x} = dx/2u</math>, that is, <math>dx = 2u\,du\,</math>, | ||
| - | {{ | + | {{Displayed math||<math>\int \ln \sqrt{x} \, dx = \int \ln u \cdot 2u \, du\,\mbox{.}</math>}} |
Then we integrate by parts. Put <math>f=2u</math> and <math>g=\ln u</math>, which gives | Then we integrate by parts. Put <math>f=2u</math> and <math>g=\ln u</math>, which gives | ||
| - | {{ | + | {{Displayed math||<math>\begin{align*}\int \ln u \cdot 2u \, du &= u^2 \ln u - \int u^2 \cdot \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}\cdot\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}\cdot\ln x</math>. | ||
</div> | </div> | ||
Version vom 08:25, 17. Sep. 2008
| 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 f and \displaystyle g are two differentiable functions then the rule for products gives
| \displaystyle D\,(\,f\cdot g) = f^{\,\prime} \cdot g + f \cdot g'\,\mbox{.} |
Now if one integrates both sides one gets
| \displaystyle f \cdot g = \int (\,f^{\,\prime} \cdot g + f \cdot g'\,)\,dx = \int f^{\,\prime} \cdot g\,dx + \int f\cdot g'\,dx |
or after re-ordering
| \displaystyle \int f^{\,\prime} \cdot g\,dx = f \cdot g - \int f \cdot g'\,dx\,\mbox{.} |
This gives us the formula for integration by parts.
Integration by parts:
| \displaystyle \int f(x)\cdot g(x)\,dx = F(x) \cdot g(x) - \int F(x) \cdot g'(x)\,dx\,\mbox{.} |
This means in practice that one integrates a product of functions by calling one factor \displaystyle f and the other \displaystyle g, and then replaces the integral \displaystyle \,\int f \cdot g\,dx\ , hopefully, by an easier integral \displaystyle \,\int F \cdot g'\,dx\,\mbox{,}\ where \displaystyle F is a primitive function of \displaystyle f and \displaystyle g' is the derivative of \displaystyle g.
It is important to note that the method does not always lead to an integral which is easier than the original. It may also be crucial how one chooses the functions \displaystyle f and \displaystyle g, as the following example shows.
Example 1
Determine the integral \displaystyle \,\int x \cdot \sin x \, dx\,.
If one chooses \displaystyle f=x and \displaystyle g=\sin x one gets \displaystyle F=x^2/2 and \displaystyle g'=\cos x, and the formula for integration by parts gives
| \displaystyle \int x \cdot \sin x \, dx = \frac{x^2}{2} \cdot \sin x - \int \frac{x^2}{2} \cdot \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 f=\sin x and \displaystyle g=x then \displaystyle F=-\cos x and \displaystyle g'=1, and
| \displaystyle \int x \cdot \sin x \, dx = - x \cdot \cos x - \int - 1 \cdot \cos x \, dx = - x\cos x + \sin x + C\,\mbox{.} |
Example 2
Determine the integral \displaystyle \ \int x^2 \cdot \ln x \, dx\,.
Put \displaystyle f=x^2 and \displaystyle g=\ln x since differentiation eliminates the logarithm when we carry out an integration by parts: \displaystyle F=x^3/3 and \displaystyle g'=1/x. This gives us that
| \displaystyle \begin{align*}\int x^2 \cdot \ln x \, dx &= \frac {x^3}{3} \cdot \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac {x^3}{3} \cdot \ln x - \frac{1}{3} \int x^2 \, dx\\[4pt] &= \frac{x^3}{3} \cdot \ln x - \frac{1}{3} \cdot \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 f=e^x and \displaystyle g=x^2, which gives that \displaystyle F=e^x and \displaystyle g'=2x, and an 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 f=e^x and \displaystyle g=2x, which gives \displaystyle F=e^x and \displaystyle g'=2
| \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 a 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 \cdot \cos x - \int e^x \cdot(-\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 \cdot e^{-x} \, dx\,\mbox{.} |
Substitute \displaystyle f=e^{-x} and \displaystyle g=2x, and integrate by parts
| \displaystyle \begin{align*}\int_{0}^{1} 2x \cdot 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 \cdot e^{-1}) - 0 + (- 2\cdot 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 variable 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 \cdot 2u \, du\,\mbox{.} |
Then we integrate by parts. Put \displaystyle f=2u and \displaystyle g=\ln u, which gives
| \displaystyle \begin{align*}\int \ln u \cdot 2u \, du &= u^2 \ln u - \int u^2 \cdot \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}\cdot\ln x.
