Integration: Integration techniques
Repeated integration by parts
We have seen how we shift calculating the antiderivative of #f(x)\cdot g'(x)# to finding the derivative of #f'(x)\cdot g(x)# when we use integration by parts. We can also find the antiderivative of #f'(x)\cdot g(x)# using integration by parts. We call this repeated integration by parts. We will now look at two situations in which repeated integration by parts yields results. First of all, we look at the situation where we find the solution directly by integrating by parts again. Here, we might have to integrate by parts more than twice.
Step-by-step guide repeated integration by parts
Step-by-step |
Example | |
Determine #\displaystyle \int f(x) \; \dd x# using integration by parts. |
Determine #\displaystyle \int x^2 \e^{2x} \dd x# | |
Step 1 |
Determine #\blue{g(x)}# and #\green{h'(x)}# such that #f(x)=\blue {g(x)} \cdot \green{h'(x)}#. |
#\blue{g(x)}=x^2# #\green{h'(x)}=\e^{2x}# |
Step 2 |
Determine #\purple{g'(x)}# and #\orange{h(x)}#. |
#\purple{g'(x)}=2x# #\orange{h(x)}=\frac{1}{2} \e^{2x}# |
Step 3 |
Determine #\int f(x) \; \dd x# with: \[\begin{array}{rcl}\displaystyle \int f(x) \; \dd x&=&\displaystyle\int \blue {g(x)} \cdot \green{h'(x)} \dd x \\ &=&\blue{g(x)} \cdot \orange{h(x)} - \displaystyle \int \purple{g'(x)} \cdot \orange{h(x)} \; \dd x \end{array}\] |
\[\blue{x^2} \cdot \orange{\frac{1}{2} \e^{2x}} - \displaystyle \int \purple{2x} \cdot \orange{\frac{1}{2} \e^{2x}} \; \dd x \] |
Step 4 |
Determine #\int \purple{g'(x)} \cdot \orange{h(x)} \; \dd x# again using integration by parts. |
#\displaystyle \int \purple{2x} \cdot \orange{\frac{1}{2} \e^{2x}} \; \dd x= 2x \cdot \frac{1}{4} \e^{2x}-\frac{1}{4} \e^{2x}# |
Step 5 |
Determine the final answer by substituting step 4 in step 3. |
# \frac{1}{2}x^2 \cdot \e^{2x}- \frac{1}{2} x \cdot \e^{2x}+\frac{1}{4} \e^{2x}+C# |
We will now look at the situation in which we find the original integral again when we repeatedly integrate by parts. Subsequently, we can determine the original integral with the aid of reduction.
Repeated integration by parts using reduction
Step-by-step |
Example | |
Determine #\displaystyle \int f(x) \; \dd x# using integration by parts. |
Determine #\displaystyle \int e^x \cdot \cos(x) \dd x# | |
Step 1 |
Determine #\blue{g(x)}# and #\green{h'(x)}# such that #f(x)=\blue {g(x)} \cdot \green{h'(x)}#. |
#\blue{g(x)}=\cos(x)# #\green{h'(x)}=\e^x# |
Step 2 |
Determine #\purple{g'(x)}# and #\orange{h(x)}#. |
#\purple{g'(x)}=-\sin(x)# #\orange{h(x)}=\e^x# |
Step 3 |
Determine #\int f(x) \; \dd x# with: \[\begin{array}{rcl}\displaystyle \int f(x) \; \dd x&=&\displaystyle\int \blue {g(x)} \cdot \green{h'(x)} \dd x \\ &=&\blue{g(x)} \cdot \orange{h(x)} - \displaystyle \int \purple{g'(x)} \cdot \orange{h(x)} \; \dd x \end{array}\] |
\[\blue{\cos(x)} \cdot \orange{\e^x} - \displaystyle \int \purple{-\sin(x)} \cdot \orange{\e^x} \; \dd x \] |
Step 4 |
Integrate #\int \purple{g'(x)} \cdot \orange{h(x)} \; \dd x# again by parts. |
#\begin{array}{rcl}&&\displaystyle \int \purple{-\sin(x)} \cdot \orange{\e^x} \; \dd x \\&=& -\sin(x) \cdot \e^x +\int \cos(x) \cdot \e^x \; \dd x\end{array}# |
Step 5 |
Write down the equation found by merging steps 3 and 4 with the original question. |
#\begin{array}{rcl}&&\displaystyle \int e^x \cdot \cos(x) \dd x\\&=&\blue{\cos(x)} \cdot \orange{\e^x} +\sin(x) \cdot \e^x \\&-&\displaystyle\int \cos(x) \cdot \e^x \; \dd x\end{array}# |
Step 6 |
Solve the obtained equation. |
#\begin{array}{rcl}&&\displaystyle \int e^x \cdot \cos(x) \dd x\\ &=&\frac{1}{2}\blue{\cos(x)} \cdot \orange{\e^x} +\frac{1}{2}\sin(x) \cdot \e^x \end{array}# |
Step 1 | We are looking for #g(x)# and #h'(x)#, such that #x^2\cdot \cos \left(x\right)=g(x) \cdot h'(x)#. In this case we choose #g(x)=x^2# and #h'(x)=\cos \left(x\right)#. |
Step 2 | Now we calculate #g'(x)# and #h(x)#. #g'(x)=2\cdot x# #h(x)=\sin \left(x\right)# |
Step 3 | According to the calculation rule for integration by parts: \[\int g(x) \cdot h'(x) \; \dd x=g(x) \cdot h(x) - \int g'(x) \cdot h(x) \; \dd x\] This gives: \[\int x^2\cdot \cos \left(x\right) \,\dd x=x^2 \cdot \sin \left(x\right) - \int 2\cdot x \cdot \sin \left(x\right) \; \dd x\] |
Step 4 | We now calcualte: #\int 2\cdot x \cdot \sin \left(x\right) \; \dd x# again with integration by parts. \[\begin{array}{rcl}\displaystyle \int 2\cdot x \cdot \sin \left(x\right) \; \dd x&=&\displaystyle 2\cdot x \cdot -\cos \left(x\right) - \int 2 \cdot -\cos \left(x\right) \\ && \phantom{xxx}\blue{\text{integration by parts with }g(x)=2\cdot x \text{ and } h'(x)=\sin \left(x\right)} \\ \displaystyle &=&\displaystyle -2\cdot x\cdot \cos \left(x\right) +2\cdot \sin \left(x\right) \\ && \phantom{xxx}\blue{\text{calculated the integral and simplified}} \end{array}\] |
Step 5 | We now substitute step 4 in step 3. This gives: \[\begin{array}{rcl}\displaystyle \int x^2\cdot \cos \left(x\right) \,\dd x&=&x^2\cdot \sin \left(x\right)-(-2\cdot x\cdot \cos \left(x\right) +2\cdot \sin \left(x\right)) + C\\ &&\phantom{xxx}\blue{\text{step 4 substituted in step 3}} \\ &=&\displaystyle x^2\cdot \sin \left(x\right)+2\cdot x\cdot \cos \left(x\right) -2\cdot \sin \left(x\right)+C \\&& \phantom{xxx}\blue{\text{expanded brackets}}\end{array}\] |
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