Integration by Substitution Integration by Substitution Integration by Substitution 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.
Basic differentiation
Basic integration methods To understand the theory:
Function composition : that a function
f
(
x
)
{\displaystyle f(x)}
f
(
u
(
x
)
)
{\displaystyle f(u(x))}
f
(
u
)
{\displaystyle f(u)}
u
(
x
)
{\displaystyle u(x)}
Chain rule 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}
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)}
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\,}
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\,}
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}}}
Let's now review the steps for integration by substitution.
Indefinite Integral
Definite Integral
1.
First identify that you have a function of a function. This skill comes with practice to identify candidates.
First identify that you have a function of a function. This skill comes with practice to identify candidates.
2.
Identify
u
{\displaystyle u}
d
u
{\displaystyle du}
Identify
u
{\displaystyle u}
d
u
{\displaystyle du}
3.
Change limits for definite integrals.
4.
Integrate using normal techniques.
Integrate using normal techniques.
5.
Substitute back the values for u for indefinite integrals.
6.
Don't forget the constant of integration for indefinite integrals.
Let's look at more examples at finding
u
{\displaystyle u}
∫
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.}
∫
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}
