Introduction to this topic
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This page is dedicated to teaching techniques for integration by substitution. For other integration methods, see other sources.
The first section introduces the theory. Next comes a demonstration of the technique; this is followed by a section listing the steps used in that demonstration. The last section is a series of clarifying examples.
Assumed Knowledge
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Theory of Integration by Substitution
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This area is covered by the Wikipedia article Integration by substitution . On this page we deal with the practical aspects.
We begin with the following as is described by the Wikipedia article
∫
a
b
f
(
g
(
x
)
)
g
′
(
x
)
d
x
=
∫
g
(
a
)
g
(
b
)
f
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}f(g(x))g'(x)\,dx=\int _{g(a)}^{g(b)}f(x)\,dx.}
This can be rewritten as
∫
f
(
u
)
d
u
.
{\displaystyle \int _{}^{}f(u)\,du.}
by setting
u
=
g
(
x
)
d
u
=
g
′
(
x
)
d
x
.
{\displaystyle u=g(x)\,\qquad du=g'(x)\,dx.}
The principle applied here is function of a function (Function composition ) and the reverse of the chain rule . This is the basis of integration by substitution.
u
{\displaystyle u}
Integration by substitution
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The objective of Integration by substitution is to substitute the integrand from an expression with variable
x
{\displaystyle x}
u
{\displaystyle u}
u
=
g
(
x
)
{\displaystyle u\,=\,g(x)}
Theory
x
{\displaystyle x}
u
{\displaystyle u}
∫
x
=
a
x
=
b
f
(
x
)
d
x
→
∫
u
=
c
u
=
d
h
(
u
)
d
u
{\displaystyle \int _{x=a}^{x=b}f(x)\,dx\,\rightarrow \,\int _{u=c}^{u=d}h(u)\,du}
u
=
g
(
x
)
{\displaystyle u\,=\,g(x)}
Steps
∫
x
=
a
x
=
b
f
(
x
)
d
x
{\displaystyle \int _{x=a}^{x=b}f(x)\,dx\,}
=
∫
x
=
a
x
=
b
f
(
x
)
d
u
d
u
d
x
{\displaystyle =\int _{x=a}^{x=b}f(x)\,{\operatorname {d} \!u \over \operatorname {d} \!u}\,dx\,}
(1)
ie
d
u
d
u
=
1
{\displaystyle {\operatorname {d} \!u \over \operatorname {d} \!u}\,=\,1}
=
∫
x
=
a
x
=
b
(
f
(
x
)
d
x
d
u
)
(
d
u
d
x
)
d
x
{\displaystyle =\int _{x=a}^{x=b}\left(f(x)\,{\operatorname {d} \!x \over \operatorname {d} \!u}\right)\left({\operatorname {d} \!u \over \operatorname {d} \!x}\right)\,dx\,}
(2)
ie
d
x
d
u
d
u
d
x
=
d
u
d
u
=
1
{\displaystyle {\operatorname {d} \!x \over \operatorname {d} \!u}{\operatorname {d} \!u \over \operatorname {d} \!x}\,=\,{\operatorname {d} \!u \over \operatorname {d} \!u}\,=\,1}
=
∫
x
=
a
x
=
b
(
f
(
x
)
d
x
d
u
)
g
′
(
x
)
d
x
{\displaystyle =\int _{x=a}^{x=b}\left(f(x)\,{\operatorname {d} \!x \over \operatorname {d} \!u}\right)g'(x)\,dx\,}
(3)
ie
d
u
d
x
=
g
′
(
x
)
{\displaystyle {\operatorname {d} \!u \over \operatorname {d} \!x}\,=\,g'(x)}
=
∫
x
=
a
x
=
b
h
(
g
(
x
)
)
g
′
(
x
)
d
x
{\displaystyle =\int _{x=a}^{x=b}h(g(x))g'(x)\,dx\,}
(4)
ie Now equate
(
f
(
x
)
d
x
d
u
)
{\displaystyle (f(x)\,{\operatorname {d} \!x \over \operatorname {d} \!u})}
h
(
g
(
x
)
)
{\displaystyle h(g(x))}
=
∫
x
=
a
x
=
b
h
(
u
)
g
′
(
x
)
d
x
{\displaystyle =\int _{x=a}^{x=b}h(u)g'(x)\,dx\,}
(5)
ie
g
(
x
)
=
u
{\displaystyle g(x)\,=\,u}
=
∫
u
=
g
(
a
)
u
=
g
(
b
)
h
(
u
)
d
u
{\displaystyle =\int _{u=g(a)}^{u=g(b)}h(u)\,du\,}
(6)
ie
d
u
=
d
u
d
x
d
x
=
g
′
(
x
)
d
x
{\displaystyle du\,=\,{\operatorname {d} \!u \over \operatorname {d} \!x}dx\,=\,g'(x)\,dx\,}
=
∫
u
=
c
u
=
d
h
(
u
)
d
u
{\displaystyle =\int _{u=c}^{u=d}h(u)\,du\,}
(7)
ie We have achieved our desired result
Procedure
Calculate
g
′
(
x
)
=
d
u
d
x
{\displaystyle g'(x)\,=\,{\operatorname {d} \!u \over \operatorname {d} \!x}}
Calculate
h
(
u
)
{\displaystyle h(u)}
f
(
x
)
d
x
d
u
=
f
(
x
)
g
′
(
x
)
{\displaystyle f(x)\,{\operatorname {d} \!x \over \operatorname {d} \!u}\,=\,{\frac {f(x)}{g'(x)}}}
make sure you express the result in terms of the variable
u
{\displaystyle u}
Calculate
c
=
g
(
a
)
{\displaystyle c\,=\,g(a)}
Calculate
d
=
g
(
b
)
{\displaystyle d\,=\,g(b)}
Technique
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Example 1
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Let us examine this integral
∫
(
x
−
3
)
10
d
x
.
{\displaystyle \int _{}^{}(x-3)^{10}\,dx.}
The inner function is
x
−
3
{\displaystyle x-3\,}
The outer function is
(
)
10
{\displaystyle (\qquad )^{10}}
Recognising this relationship we then move onto the following set of steps to process the inner function
x
−
3
{\displaystyle x-3}
1
{\displaystyle 1}
u
=
x
−
3
d
u
d
x
=
1
d
u
=
1
×
d
x
d
u
=
d
x
{\displaystyle u=x-3\,\qquad {\frac {du}{dx}}=1\,\qquad du=1\times dx\,\qquad du=dx\,}
Now we substitute
u
{\displaystyle u}
d
u
{\displaystyle du}
∫
(
x
−
3
)
10
d
x
.
=
∫
(
u
)
10
d
u
.
{\displaystyle \int _{}^{}(x-3)^{10}\,dx.\qquad =\int _{}^{}(u)^{10}\,du.}
Then apply standard integral technique
u
n
+
1
n
+
1
+
c
u
11
11
+
c
{\displaystyle {\frac {u^{n+1}}{n+1}}+c\,\qquad {\frac {u^{11}}{11}}+c\,}
And finally we substitute the value of
u
{\displaystyle u}
(
x
−
3
)
11
11
+
c
{\displaystyle {\frac {(x-3)^{11}}{11}}+c\,}
Example 2
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Let us examine this integral
∫
x
9
+
x
2
d
x
.
{\displaystyle \int {\frac {x}{\sqrt {9+x^{2}}}}\,dx.}
We can first rearrange the fraction to make it more familiar.
∫
x
(
9
+
x
2
)
−
1
2
d
x
.
{\displaystyle \int x(9+x^{2})^{-{\frac {1}{2}}}\,dx.}
The inner function is
9
+
x
2
{\displaystyle 9+x^{2}\,}
The outer function is
x
(
)
−
1
2
{\displaystyle x(\qquad )^{-{\frac {1}{2}}}}
Next we assign
u
{\displaystyle u}
d
u
{\displaystyle du}
u
=
9
+
x
2
d
u
d
x
=
2
x
{\displaystyle u=9+x^{2}\,\qquad {\frac {du}{dx}}=2x\,}
But we have a problem!
d
u
{\displaystyle du\,}
x
{\displaystyle x\,}
d
u
{\displaystyle du\,}
d
u
d
x
=
2
x
d
u
2
=
x
d
x
1
2
d
u
=
x
d
x
{\displaystyle {\frac {du}{dx}}=2x\,\qquad {\frac {du}{2}}=xdx\,\qquad {\frac {1}{2}}du=xdx\,}
Now we can substitute
u
{\displaystyle u}
d
u
{\displaystyle du}
∫
x
(
9
+
x
2
)
−
1
2
d
x
.
1
2
∫
(
u
)
−
1
2
d
u
.
{\displaystyle \int x(9+x^{2})^{-{\frac {1}{2}}}\,dx.\qquad {\frac {1}{2}}\int (u)^{-{\frac {1}{2}}}\,du.}
Study the above substitution carefully. We moved the fractional component of du to the front as it represents a constant.
Now apply standard integral technique
u
n
+
1
n
+
1
+
c
1
2
.
u
1
2
1
2
+
c
{\displaystyle {\frac {u^{n+1}}{n+1}}+c\,\qquad {\frac {1}{2}}.{\frac {u^{\frac {1}{2}}}{\frac {1}{2}}}+c\,}
Cleaning up this expression we have
u
1
2
+
c
{\displaystyle u^{\frac {1}{2}}+c\,}
And finally we substitute the value of u back into the equation
(
9
+
x
2
)
1
2
+
c
9
+
x
2
+
c
{\displaystyle (9+x^{2})^{\frac {1}{2}}+c\,\qquad {\sqrt {9+x^{2}}}+c\,}
The Definite Integral
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Consider the definite integral
∫
0
2
x
cos
(
x
2
+
1
)
d
x
{\displaystyle \int _{0}^{2}x\cos(x^{2}+1)\,dx}
By using the substitution
u
(
x
)
=
x
2
+
1
d
u
d
x
=
2
x
d
u
2
=
x
d
x
{\displaystyle u(x)=x^{2}+1\qquad {\frac {du}{dx}}=2x\,\qquad {\frac {du}{2}}=x\,dx}
Now because we have limits, we need to change them with respect to
u
{\displaystyle u}
∫
x
=
0
x
=
2
x
cos
(
x
2
+
1
)
d
x
{\displaystyle \int _{x=0}^{x=2}x\cos(x^{2}+1)\,dx}
u
(
0
)
=
(
0
2
+
1
)
=
1
u
(
2
)
=
(
2
2
+
1
)
=
5
{\displaystyle {\begin{aligned}&{}u(0)=(0^{2}+1)=1\\&{}u(2)=(2^{2}+1)=5\end{aligned}}}
Now we have a new definite integral to solve
=
1
2
∫
u
=
1
u
=
5
cos
u
d
u
=
1
2
(
sin
5
−
sin
1
)
.
{\displaystyle {\begin{aligned}&{}={\frac {1}{2}}\int _{u=1}^{u=5}\cos \,u\,du\\&{}={\frac {1}{2}}(\sin 5-\sin 1).\end{aligned}}}
The Steps We Applied
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Finding u
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Let's look at more examples at finding
u
{\displaystyle u}
Example 1
edit
∫
0
1
x
cosh
4
−
x
2
4
−
x
2
d
x
.
{\displaystyle \int _{0}^{1}{\frac {x\cosh {\sqrt {4-x^{2}}}}{\sqrt {4-x^{2}}}}\,dx.}
u
{\displaystyle u\,}
d
u
d
x
{\displaystyle {\frac {du}{dx}}\,}
L
i
m
i
t
s
{\displaystyle Limits\,}
u
I
n
t
e
g
r
a
l
{\displaystyle u\,Integral\,}
u
=
4
−
x
2
{\displaystyle u={\sqrt {4-x^{2}}}}
d
u
d
x
=
1
2
(
4
−
x
2
)
−
1
2
×
−
2
x
{\displaystyle {\frac {du}{dx}}={\frac {1}{2}}(4-x^{2})^{\frac {-1}{2}}\times -2x\,}
−
∫
x
=
0
x
=
1
cosh
u
d
u
.
{\displaystyle -\int _{x=0}^{x=1}\cosh \,u\,du.}
−
∫
u
=
2
u
=
3
cosh
u
d
u
{\displaystyle -\int _{u=2}^{u={\sqrt {3}}}\cosh \,u\,du}
=
−
x
(
4
−
x
2
)
−
1
2
{\displaystyle =-x(4-x^{2})^{\frac {-1}{2}}}
u
=
4
−
1
2
=
3
{\displaystyle u={\sqrt {4-1^{2}}}={\sqrt {3}}}
=
−
(
sinh
2
−
sinh
3
)
{\displaystyle =-(\sinh \,2-\sinh \,{\sqrt {3}})}
=
−
x
4
−
x
2
{\displaystyle ={\frac {-x}{\sqrt {4-x^{2}}}}}
u
=
4
−
0
2
=
2
{\displaystyle u={\sqrt {4-0^{2}}}=2}
=
−
sinh
2
+
sinh
3
{\displaystyle =-\sinh \,2+\sinh \,{\sqrt {3}}}
d
u
=
−
x
4
−
x
2
d
x
.
{\displaystyle du={\frac {-x}{\sqrt {4-x^{2}}}}\,dx.}
Example 2
edit
∫
1
x
(
4
+
ln
2
x
)
d
x
.
{\displaystyle \int {\frac {1}{x(4+\ln ^{2}x)}}\,dx.}
Here we first perform the substitution
u
=
ln
x
{\displaystyle u=\ln x}
u
=
ln
x
⇒
d
u
=
1
x
d
x
⇒
d
x
=
e
u
d
u
{\displaystyle u=\ln x\Rightarrow du={\frac {1}{x}}dx\Rightarrow dx=e^{u}du}
With this, we get
∫
1
x
(
4
+
ln
2
x
)
d
x
=
∫
e
u
e
u
(
4
+
u
2
)
d
u
=
∫
1
4
+
u
2
d
u
=
{\displaystyle \int {\frac {1}{x(4+\ln ^{2}x)}}\,dx=\int {\frac {e^{u}}{e^{u}(4+u^{2})}}\,du=\int {\frac {1}{4+u^{2}}}\,du=}
=
1
2
arctan
(
u
/
2
)
+
c
=
1
2
arctan
(
ln
x
2
)
+
c
{\displaystyle ={\frac {1}{2}}\arctan(u/2)+c={\frac {1}{2}}\arctan \left({\frac {\ln x}{2}}\right)+c}
