Integration by Substitution
This method is also called the u-substitution or the reverse of chain rule of derivation.
The chain rule except being useful in derivation is also in integration:
If we have two functions $ \displaystyle f(x)$ and $ \displaystyle g(x)$ then the derivative of their composite function is:$ \displaystyle (f\circ g{)}'(x)={f}'(g(x)){g}'(x)$.
How it helps on integration of a function?
Integrating by substituting $ \displaystyle u=ax+b$.
Let’s observe.
Example 1: How to find the integral of $ \displaystyle \int{{{{{(x+10)}}^{4}}}}dx$?
We know that if the term 10 misses then it’s easy to find the integral by using the power rule of integration.
$ \displaystyle \int{{{{x}^{4}}}}=\frac{{{{x}^{5}}}}{5}+C$
What would happen if we apply the power rule to our example?
$ \displaystyle \int{{{{{(x+10)}}^{4}}}}dx=\frac{{{{{(x+10)}}^{5}}}}{5}+C$
The result is correct but it’s not always true!
Let’s see what happens if we use substitution of the term $ \displaystyle u=x+10$.
The variable becomes $ \displaystyle {{(x+10)}^{4}}={{u}^{4}}$
Now the variable of integration got changed into u, so we also need to substitute the term dx with du. Since both du and dx are differentials then: $ \displaystyle du=(\frac{{du}}{{dx}})dx$
To find du in relation to dx we find the derivative of $ \displaystyle u=x+10$.
$ \displaystyle {u}’=(x+10{)}’$
$ \displaystyle du=1\cdot dx$
$ \displaystyle du=dx$
Substituting and finding the integral will give us:
$ \displaystyle \int{{{{{(x+10)}}^{4}}}}dx=\int{{{{u}^{4}}}}du=\frac{{{{u}^{5}}}}{5}+C$
Now we can substitute back $ \displaystyle u=x+10$ and get:
$ \displaystyle \int{{{{{(x+10)}}^{4}}}}dx=\frac{{{{{(x+10)}}^{5}}}}{5}+C$
We see that it´s the same answer as when applying the power rule but let’s see another example when this is not true.
Example 2: How to find the integral of $ \displaystyle \int{{{{{(3x+5)}}^{2}}}}dx$?
Substituting the term $ \displaystyle u=ax+b$
$ \displaystyle u=3x+5$
$ \displaystyle du=3dx$
$ \displaystyle dx=\frac{1}{3}du$
Finding the integral by substituting will give us:
$ \displaystyle \int{{{{{(3x+5)}}^{2}}}}dx=\int{{{{u}^{2}}}}\frac{1}{3}du$
$ \displaystyle \int{{{{u}^{2}}}}\frac{1}{3}du=\frac{1}{3}\int{{{{u}^{2}}}}du=\frac{1}{3}\cdot \frac{{{{u}^{3}}}}{3}+C=\frac{{{{u}^{3}}}}{9}+C$
Now we can substitute back the $ \displaystyle u=3x+5$
$ \displaystyle \int{{{{{(3x+5)}}^{2}}}}dx=\frac{{{{{(3x+5)}}^{3}}}}{9}+C$
But if we used directly the power rule we would get:$ \displaystyle \int{{{{{(3x+5)}}^{2}}}}dx=\frac{{{{{(3x+5)}}^{3}}}}{3}+C$ which is not the same because the coefficient before x makes the difference so be careful!
Use the u-substitution rule to be correct.
Example 3: Find the integral $ \displaystyle \int{{\cos (6x+2)dx}}$ using the u-substitution method?
The obvious substitution here will be $ \displaystyle u=6x+2$
Then we will find $ \displaystyle du=(\frac{{du}}{{dx}})dx$
$ \displaystyle du=6dx$
$ \displaystyle dx=\frac{1}{6}du$
Finding the integral by substituting will give us:
$ \displaystyle \int{{\cos (6x+2)dx}}=\int{{\frac{1}{6}\cos udu}}$
$ \displaystyle \int{{\frac{1}{6}\cos udu}}=\frac{1}{6}\sin u+C$
Now we can substitute back the $ \displaystyle u=6x+2$
$ \displaystyle \int{{\cos (6x+2)dx}}=\frac{{\sin (6x+2)}}{6}+C$
Integrating the product of two functions
We mention that u-substitution does to integration the same that does the chain rule to differentiation. So how we integrate composite functions using substitution rule?
$ \displaystyle \int{{f(g(x)){g}'(x)dx=\int{{f(u)du}}}}$
Where $ \displaystyle u=g(x)$ and $ \displaystyle du={g}'(x)dx$
Example 4: How to find the integral of $ \displaystyle \int{{2x\cos {{{(x)}}^{2}}}}dx$?
In this case it´s obvious that we will substitute $ \displaystyle u={{x}^{2}}$ and then $ \displaystyle du=2xdx$
Now all its left is substitute and integrate
$ \displaystyle \int{{2x\cos {{{(x)}}^{2}}}}dx=\int{{\cos udu=\sin u+C}}$
Now we substitute back $ \displaystyle u={{x}^{2}}$
$ \displaystyle \int{{2x\cos {{{(x)}}^{2}}}}dx=\sin ({{x}^{2}})+C$
This was an easy example but what if things are not this simple?
Example 5: How to find this integral $ \displaystyle \int{{\sin ({{x}^{2}}}})12xdx$?
If we see it carefully this example is similar to example 4, with the only difference the coefficient 12.
Instead of 2x we have 12x.
If you recall the constant coeficcient rule, then we can do this:
$ \displaystyle \int \sin ({{x}^{2}})12xdx=6\int{{\sin ({{x}^{2}}}})2xdx$
Now we got under the integral the same expression as above, which we know how to integrate by using the u-substitution method.
$ \displaystyle 6\int{{\sin ({{x}^{2}}}})2xdx=6\int{{\sin udu}}$
$ \displaystyle 6\int{{\sin udu}}=-6\cos u+C$
Substituting back the u we get:
$ \displaystyle \int \sin ({{x}^{2}})12xdx=-6\cos {{x}^{2}}+C$
Example 6: Find the integral $ \displaystyle \int{{{{x}^{3}}}}\cos {{x}^{4}}dx$.
If we substitute $ \displaystyle {{x}^{4}}=u$ then $ \displaystyle 4{{x}^{3}}dx=du$
If we see carefully in our integral we have only $ \displaystyle {{{x}^{3}}}$ but if we multiply and divide with 4 and use the constant coeficcient rule then we can arrange things the way it will help us.
$ \displaystyle \int{{\frac{4}{4}}}{{x}^{3}}\cos {{x}^{4}}dx=\frac{1}{4}\int{4}{{x}^{3}}\cos {{x}^{4}}dx$
Now we can use the u-substitution rule:
$ \displaystyle \frac{1}{4}\int{4}{{x}^{3}}\cos {{x}^{4}}dx=\frac{1}{4}\int{{\cos udu}}$
$ \displaystyle \frac{1}{4}\int{{\cos udu}}=\frac{1}{4}\sin u+C$
Substituting back the u we get:
$ \displaystyle \int{{{{x}^{3}}}}\cos {{x}^{4}}dx=\frac{1}{4}\sin {{x}^{4}}+C$
Integrating the quotient of two functions
When we find the derivative of a quotient we use the quotient rule, but in this case we don´t have a general rule so basically the things are a little bit more difficult.
Example 7: Find the integral $ \displaystyle \int{{\frac{{{{x}^{2}}}}{{{{x}^{3}}+3}}}}dx$.
If we substitute $ \displaystyle u={{x}^{3}}+3$ then $ \displaystyle du=3{{x}^{2}}dx$
But as we can see we only have $ \displaystyle {{{x}^{2}}}$,but if we multiply and divide by 3 and use the constant coefficient rule then we can arrange things the way it will help us.
$ \displaystyle \int{{\frac{{3({{x}^{2}})}}{{3({{x}^{3}}+3)}}}}dx=\frac{1}{3}\int{{\frac{{3{{x}^{2}}}}{{{{x}^{3}}+3}}}}dx$
Now we can use the u-substitution rule:
$ \displaystyle \frac{1}{3}\int{{\frac{{3{{x}^{2}}}}{{{{x}^{3}}+3}}}}dx=\frac{1}{3}\int{{\frac{1}{u}}}du$
$ \displaystyle \frac{1}{3}\int{{\frac{1}{u}}}du=\frac{1}{3}\ln u+C$
Substituting back the u we get:
$ \displaystyle \int{{\frac{{{{x}^{2}}}}{{{{x}^{3}}+3}}}}dx=\frac{1}{3}\ln ({{x}^{3}}+3)+C$
Let’s see another example when the denominator is a radical.
Example 8: Find the integral $ \displaystyle \int{{\frac{{2x}}{{\sqrt[3]{{{{x}^{2}}+1}}}}dx}}$.
If we substitute $ \displaystyle u={{x}^{2}}+1$ then $ \displaystyle du=2xdx$
Now we can use the u-substitution rule:
$ \displaystyle \int{{\frac{{2x}}{{\sqrt[3]{{{{x}^{2}}+1}}}}dx=\int{{\frac{{du}}{{\sqrt[3]{u}}}}}}}$
To solve the integral we need to use the power rule of integration .
$ \displaystyle \int{{\frac{{du}}{{\sqrt[3]{u}}}}}=\int{{{{u}^{{\frac{1}{3}}}}}}du=\frac{{{{u}^{{\frac{4}{3}}}}}}{{\frac{4}{3}}}+C=\frac{3}{4}\sqrt[3]{{{{u}^{4}}}}+C$
Substituting back the u we get:
$ \displaystyle \int{{\frac{{2x}}{{\sqrt[3]{{{{x}^{2}}+1}}}}dx}}=\frac{3}{4}\sqrt[3]{{{{{({{x}^{2}}+1)}}^{4}}}}+C$
Based on the examples we observed above we got on conclusion:
$ \displaystyle \int{{\cos (ax+b)dx=\frac{1}{a}}}\sin (ax+b)+C$
$ \displaystyle \int{{\sin (ax+b)dx=-\frac{1}{a}}}\cos (ax+b)+C$
$ \displaystyle \int{{\frac{1}{{ax+b}}}}dx=\frac{1}{a}\ln (ax+b)+C$