Integration by Substitution
In this topic we shall see an important method for evaluating many complicated integrals.
Substitution for integrals corresponds to the chain rule for derivatives.
Suppose that F (u) is an antiderivative of f (u):
\[\int {f\left( u \right)du} = F\left( u \right) + C.\]
Assuming that u = u (x) is a differentiable function and using the chain rule, we have
\[\frac{d}{{dx}}F\left( {u\left( x \right)} \right) = F^\prime\left( {u\left( x \right)} \right)u^\prime\left( x \right) = f\left( {u\left( x \right)} \right)u^\prime\left( x \right).\]
Integrating both sides gives
\[\int {f\left( {u\left( x \right)} \right)u^\prime\left( x \right)dx} = F\left( {u\left( x \right)} \right) + C.\]
Hence
\[\int {{f\left( {u\left( x \right)} \right)}{u^\prime\left( x \right)}dx} = \int {f\left( u \right)du},\;\;\text{where}\;\;u = u\left( x \right).\]
This is the substitution rule formula for indefinite integrals.
Note that the integral on the left is expressed in terms of the variable \(x.\) The integral on the right is in terms of \(u.\)
The substitution method (also called \(u-\)substitution) is used when an integral contains some function and its derivative. In this case, we can set \(u\) equal to the function and rewrite the integral in terms of the new variable \(u.\) This makes the integral easier to solve.
Do not forget to express the final answer in terms of the original variable \(x!\)
Solved Problems
Example 1.
Compute the integral \[\int {{e^{\frac{x}{2}}}dx}.\]
Solution.
Let \(u = \frac{x}{2}.\) Then
\[du = \frac{{dx}}{2},\;\; \Rightarrow dx = 2du.\]
Now we can easily integrate:
\[\int {{e^{\frac{x}{2}}}dx} = \int {{e^u} \cdot 2du} = 2\int {{e^u}du} = 2{e^u} + C = 2{e^{\frac{x}{2}}} + C.\]
Example 2.
Find the integral \[\int {{{\left( {3x + 2} \right)}^5}dx}.\]
Solution.
We make the substitution \(u = 3x + 2.\) Then
\[du = d\left( {3x + 2} \right) = 3dx.\]
The differential \(dx\) is given by
\[dx = \frac{{du}}{3}.\]
Plug all this in the integral:
\[\int {{{\left( {3x + 2} \right)}^5}dx} = \int {{u^5}\frac{{du}}{3}} = \frac{1}{3}\int {{u^5}du} = \frac{1}{3} \cdot \frac{{{u^6}}}{6} + C = \frac{{{u^6}}}{{18}} + C = \frac{{{{\left( {3x + 2} \right)}^6}}}{{18}} + C.\]
Example 3.
Find the integral \[\int {\frac{{dx}}{{\sqrt {1 + 4x} }}}.\]
Solution.
We can try to use the substitution \(u = 1 + 4x.\) Hence
\[du = d\left( {1 + 4x} \right) = 4dx,\]
so
\[dx = \frac{{du}}{4}.\]
This yields
\[\int {\frac{{dx}}{{\sqrt {1 + 4x} }}} = \int {\frac{{\frac{{du}}{4}}}{{\sqrt u }}} = \frac{1}{4}\int {\frac{{du}}{{\sqrt u }}} = \frac{1}{4}\int {{u^{ - \frac{1}{2}}}du} = \frac{1}{4} \cdot \frac{{{u^{\frac{1}{2}}}}}{{\frac{1}{2}}} + C = \frac{1}{4} \cdot 2{u^{\frac{1}{2}}} + C = \frac{{{u^{\frac{1}{2}}}}}{2} + C = \frac{{\sqrt u }}{2} + C = \frac{{\sqrt {1 + 4x} }}{2} + C.\]
Example 4.
Evaluate the integral \[\int {\frac{{xdx}}{{\sqrt {1 + {x^2}} }}}.\]
Solution.
Let \(u = 1 + {x^2}.\) Then
\[du = d\left( {1 + {x^2}} \right) = 2xdx.\]
We see that
\[xdx = \frac{{du}}{2}.\]
Hence
\[\int {\frac{{xdx}}{{\sqrt {1 + {x^2}} }}} = \int {\frac{{\frac{{du}}{2}}}{{\sqrt u }}} = \int {\frac{{du}}{{2\sqrt u }}} = \sqrt u + C = \sqrt {1 + {x^2}} + C.\]
Example 5.
Calculate the integral \[\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}}.\]
Solution.
Let \(u = \frac{x}{a}.\) Then \(x = au,\) \(dx = adu.\) Hence, the integral is
\[\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = \int {\frac{{adu}}{{\sqrt {{a^2} - {{\left( {au} \right)}^2}} }}} = \int {\frac{{adu}}{{\sqrt {{a^2}\left( {1 - {u^2}} \right)} }}} = \int {\frac{{\cancel{a}du}}{{\cancel{a}\sqrt {1 - {u^2}} }}} = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} = \arcsin u + C = \arcsin \frac{x}{a} + C.\]
Example 6.
Evaluate the integral \[{\int {{\frac{{{x^2}}}{{{x^3} + 1}}}dx} }\] using an appropriate substitution.
Solution.
We try the substitution \(u = {x^3} + 1.\)
Calculate the differential \(du:\)
\[du = d\left( {{x^3} + 1} \right) = 3{x^2}dx.\]
We see from the last expression that
\[{x^2}dx = \frac{{du}}{3},\]
so we can rewrite the integral in terms of the new variable \(u:\)
\[I = \int {\frac{{{x^2}}}{{{x^3} + 1}}dx} = \int {\frac{{\frac{{du}}{3}}}{u}} = \int {\frac{{du}}{{3u}}} .\]
Now we can easily evaluate this integral:
\[I = \int {\frac{{du}}{{3u}}} = \frac{1}{3}\int {\frac{{du}}{u}} = {\frac{1}{3}\ln \left| u \right|} + C.\]
Express the result in terms of the variable \(x:\)
\[I = \frac{1}{3}\ln \left| u \right| + C = \frac{1}{3}\ln \left| {{x^3} + 1} \right| + C.\]
Example 7.
Find the integral \[\int {\sqrt[3]{{1 - 3x}}dx}.\]
Solution.
We use the substitution \(u = 1 - 3x.\) Then
\[du = d\left( {1 - 3x} \right) = - 3dx\]
and
\[dx = - \frac{{du}}{3}.\]
After substitution we get
\[\int {\sqrt[3]{{1 - 3x}}dx} = \int {\sqrt[3]{u}\left( { - \frac{{du}}{3}} \right)} = - \frac{1}{3}\int {\sqrt[3]{u}du} = - \frac{1}{3}\int {{u^{\frac{1}{3}}}du} = - \frac{1}{3} \cdot \frac{{{u^{\frac{4}{3}}}}}{{\frac{4}{3}}} + C = - \frac{1}{3} \cdot \frac{{3{u^{\frac{4}{3}}}}}{4} + C = - \frac{{{u^{\frac{4}{3}}}}}{4} + C = - \frac{{\sqrt[3]{{{u^4}}}}}{4} + C = - \frac{{\sqrt[3]{{{{\left( {1 - 3x} \right)}^4}}}}}{4} + C.\]
Example 8.
Find the integral \[\int {{\frac{{x + 1}}{{{x^2} + 2x - 5}}} dx}.\]
Solution.
We make the substitution \(u = {x^2} + 2x - 5.\) Then
\[du = 2xdx + 2dx = 2\left( {x + 1} \right)dx\]
or
\[\left( {x + 1} \right)dx = \frac{{du}}{2}.\]
The integral is easy to calculate with the new variable:
\[\int {\frac{{x + 1}}{{{x^2} + 2x - 5}}dx} = \int {\frac{{\frac{{du}}{2}}}{u}} = \frac{1}{2}\int {\frac{{du}}{u}} = \frac{1}{2}\ln \left| u \right| + C = \frac{1}{2}\ln \left| {{x^2} + 2x - 5} \right| + C.\]
See more problems on Page 2.
