Integration: Integration techniques
Substitution method
We have seen the chain rule when we studied differentiation. This rule states #\left(g\left(h\left(x\right)\right)\right)'=g'(h(x))\cdot h'(x)#. Reversed, we can use this rule for integration.
If we can write #f(x)=\blue{g\bigl(}\green{h(x)}\blue{\bigr)} \cdot \purple{h'(x)}# and #\orange G# is an antiderivative of #\blue g#, then:
\[\begin{array}{rcl}\displaystyle \int f(x) \; \dd x &=&\displaystyle \int \blue {g\bigl(}\green{h(x)}\blue{\bigr)} \cdot \purple{ h'(x)} \; \dd x\\ &=&
\displaystyle \int \blue g\bigl(\green{h(x)}\bigr) \; \dd\bigl(\green{h(x)}\bigr) \\
&=&
\displaystyle \int \blue g\bigl(\green{u}\bigr) \; \dd \green{u} \\
&=& \orange G(\green u )
\end{array}\]
We have used the substitution #\green{u}=\green{h(x)}# here.
Example
#\begin{array}{rcl}&&\displaystyle \int 10x (x^2+1)^4 \; \dd x\\ &=&\displaystyle \int \blue{5\bigl(}\green{x^2+1}\blue{\bigr)^4} \cdot \purple{2x} \; \dd x\\ &=&
\displaystyle \int \blue{5\bigl(}\green{x^2+1}\blue{\bigr)^4}\; \dd\bigl(\green{x^2+1}\bigr) \\
&=&
\displaystyle\int \blue{5\bigl(}\green{u}\blue{\bigr)^4}\;\dd \green{u} \\
&=& \green u^5 \\ &=& (x^2+1)^5
\end{array}#
Step-by-step plan |
Example |
|
Determine #\displaystyle \int f(x) \; \dd x# using the substitution method. |
Calculate #\int 10x (x^2 +1)^4 \;\dd x#. |
|
Step 1 |
Determine the functions #\blue{g(x)}# and #\green{h(x)}#, such that #f(x)=\blue {g\bigl(}\green{h(x)}\blue{\bigr)}\cdot\purple{h'(x)} #, in which the antiderivative #\orange G# of #\blue{g(x)}# is known. |
#\begin{array}{rcl} |
Step 2 |
Rewrite the integral to the form: \[\int \blue {g\bigl(}\green{h(x)}\blue{\bigr)} \cdot \purple{ h'(x)} \; \dd x \] |
#\begin{array}{rcl} |
Step 3 |
Use #\purple{h'(x)} \; \dd x= \dd\bigl(\green {h(x)}\bigr)# to write: \[\int \blue {g\bigl(}\green{h(x)}\blue{\bigr)} \cdot \dd(\green{h(x)} \] |
#\begin{array}{rcl}&=& \displaystyle\int 5(x^2 +1)^4 \; \dd\bigl(x^2+1\bigr)\end{array}# |
Step 4 |
Substitute #\green{h(x)}=\green{u}#. |
#\begin{array}{rcl}&=&\displaystyle\int 5u^4 \;\dd u\end{array}# |
Step 5 |
Calculate the integral. |
#\begin{array}{rcl}&=&\displaystyle u^5\end{array}# |
Step 6 |
Substitute #\green{u}=\green{h(x)}# back into the answer. |
#\begin{array}{rcl}&=&\displaystyle (x^2+1)^5\end{array}# |
We apply the substitution method with #g(x)={{x^4}\over{9}}# and #h(x)=9\cdot x-2#, because in that case #g(h(x)) \cdot h'(x)=\left(9\cdot x-2\right)^4#. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \left(9\cdot x-2\right)^4 \,\dd x&=& \displaystyle \int {{\left(9\cdot x-2\right)^4}\over{9}} \cdot 9 \, \dd x \\ &&\phantom{xxx}\blue{\text{step 2: rewritten in the form }\int g(h(x)) \cdot h'(x) \, \dd x \text{ with } h'(x)=9} \\ &=& \displaystyle \int {{\left(9\cdot x-2\right)^4}\over{9}} \, \dd(9\cdot x-2) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(x)=\dd (h(x))} \\ &=& \displaystyle \int {{u^4}\over{9}} \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }9\cdot x-2=u} \\ &=& \displaystyle {{u^5}\over{45}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\left(9\cdot x-2\right)^5}\over{45}} +C \\ &&\phantom{xxx}\blue{\text{step 6: substituted }u=9\cdot x-2}
\end{array}\]
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