- Let's calculate the indefinite integral \(\int x\sqrt{1+x^2}\,\mathrm dx\).
- So far, we could integrate noninteger powers of \(x\) only using the power rule.
- We can vastly increase the number of functions using the
*substitution rule*, which for antiderivatives what the chain rule is for derivatives. - We want to find the antiderivative of \(f(x)=x\sqrt{1+x^2}\), that is a function \(F(x)\) so that \(F'(x)=f(x)\).
- When using the substitution rule, we want to change variables from \(x\) to \(u(x)\).
- That is, we are searching for functions \(F(u)\) and \(u(x)\) such that \[ \frac{\mathrm dF(u(x))}{\mathrm dx}=f(x). \]
- Using the chain rule, we see that we need \[ F'(u(x))u'(x)=f(x)=x\sqrt{1+x^2} \]
- Upon inspection, we see that we should have \(u(x)=1+x^2\), and \(F(u)=ku^{3/2}\), where \(k\) is a constant.
- To find \(k\): we need \[ F'(u(x))u'(x)=\frac32k\sqrt{1+x^2}2x=x\sqrt{1+x^2}, \] that is \(k=\frac13\).
- This means that we get \(\int x\sqrt{1+x^2}\,\mathrm dx=F(u(x))=\frac13(1+x^2)^{3/2}\).

- In the previous example, we have searched for the antiderivative at once.
- It is possible, to change variables in an indefinite integral, without immediately evaluating it.
- For this, consider the following equalities. \[ \int F'(u(x))u'(x)\,\mathrm dx=F(u(x))+C=F(u)+C=\int F'(u)\,\mathrm du. \]
*Substitution rule.*Suppose that \(I\) is an interval, \(u\) is a function differentiable on \(I\), and \(f\) is a function continuous on \(I\). Then we have the following formula. \[ \int f(u(x))u'(x)\,\mathrm dx=\int f(u)\,\mathrm du \]- The following is an alternative way of expressing the formula. \[ \int f(u(x))\frac{\mathrm du}{\mathrm dx}\mathrm dx=\int f(u)\mathrm du. \]
- Just as in the case of the chain rule, it looks like as if we were dealing with fractionals with differentials in them, and we could simplify. This is not true, but the similarity can help to remember the formula.
- Exercises. 5.5: 2, 6, 10, 12, 18, 20, 24, 26, 30, 34, 36, 42, 44, 46, 48

- Consider now a definite integral \(\int_a^bF'(u(x))u'(x)\,\mathrm dx\).
- Since we have \(\int F'(u(x))u'(x)\,\mathrm dx=F(u(x))+C\), we get \[ \int_a^bF'(u(x))\,\mathrm dx=F(u(x=b))-F(u(x=a)). \]
- But note that as \(\int F'(u)\,\mathrm du=F(u)\), we also have \[ \int_u(x=a)^u(x=b)F'(u)\,\mathrm du=F(u=u(x=b))-F(u=u(x=a)). \]
*Substitution rule for definite integrals.*Suppose that \(u\) is a differentiable function on an open interval containing \([a,b]\) and \(f\) is a continuous function on \([a,b]\). Then we have \[ \int_a^bf(u(x))u'(x)\,\mathrm dx=\int_{u(x=a)}^{u(x=b)}f(u)\,\mathrm du. \]- That is, we can substitute in the boundaries before evaluating the integrals.
*Warning.*When you use the substitution rule for definite integrals, never forget to change the boundaries.- Exercises. 5.5: 54, 56, 62, 78, 84