1 Integration by Substitution
1.1 Indefinite Integral
\(\int f(x) \mathrm{d}x = F(x) + C\) on \(I\)
The above equation is called an indefinite integral. It is indefinite because no limits of integration \(a\) and \(b\) have been specified.
\(f(x):\) is a continous function on an interval \(I\)
\(F(x):\) is an antiderivative of \(f\) on an interval \(I\).
In practice, mathematicians rarely specify the interval on which an indefinite interval applies. It's assumed that the interval of \(F\) is same as the interval of the domain of \(f\).
1.2 Some Indefinite integrals
For every rational number \(r \neq -1\), \(\int x^r \mathrm{d}x = \dfrac{x^{r+1}}{r+1} + C\)
\(\int \sin x \mathrm{d}x = -\cos x + c\)
\(\int \cos x \mathrm{d}x = \sin x + c\)
\(\int \sec^2 x \mathrm{d}x = \tan x + c\)
\(\int \csc^2 x \mathrm{d}x = -\cot x + c\)
\(\int \sec x \tan x \mathrm{d}x = \sec x + c\)
\(\int \csc x \cot x \mathrm{d}x = -\csc x + c\)
1.3 More generalization of earlier theorems
\(\int cf(x) \mathrm{d}x = c \int f(x) \mathrm{d}x\)
\(\int (f(x) + g(x)) \mathrm{d}x = \int f(x) \mathrm{d}x + \int g(x) \mathrm{d}x\)
\(\int (f(x) - g(x)) \mathrm{d}x = \int f(x) \mathrm{d}x - \int g(x) \mathrm{d}x\)
1.4 Integration by Substitution
We have seen Chain rule before:
If \(\dfrac{d}{du} (F(u)) = f(u)\) for all \(u \in I\) then
\(\dfrac{d}{dx} (F(g(x))) = f(g(x)).g'(x)\) for all \(x \in J\)
Reversing the two derivatives in the above statement, we get the following statement about indefinite integrals:
If \(\int f(u) \mathrm{d}u = F(u) + C\) on \(I\) then
\(\int f(g(x))g'(x) \mathrm{d}x = F(g(x)) + C\) on \(J\)
1.4.1 Symbolic Procedure
You are given integral of the form \(\int f(g(x)). g'(x) \mathrm{d}x\)
- Step one: Replace \(g(x)\) with \(u\) and \(g'(x) \mathrm{d}x\) with \(\mathrm{d}u\). So the original integrand will be coverted to
\(\int f(u) \mathrm{d}u\)
- Step two: Evaluated the above integral to get the solution. Replace \(u\) in the solution with \(g(x)\) to get \(F(g(x)) + c\).
1.4.2 Example
\(\int 3x^2(x^3 + 2)^4 \mathrm{d}x\)
Let \(u = x^3 + 2\), so that \(du = 3x^2 dx\). Then we have
\(\int 3x^2(x^3 + 2)^4 \mathrm{d}x = \int u^4 du = \dfrac{u^5}{5} + C = \dfrac{(x^3+2)^5}{5} + C\)
Note that although Leibniz derivative notation \(\dfrac{du}{dx}\) is not really a fraction, the chain rule says that in some ways it acts like one.
1.5 Theorem 5.5.3
Suppose \(f\) is continous on an interval \(I\), \(g\) is differentiable on an interval \(J\), \(g'\) is continous on \(J\), and for every \(x \in J\), \(g(x) \in I\). Then for any \(a\) and \(b\) in \(J\)
\(\int_a^b f(g(x)).g'(x) \mathrm{d}x = \int_{g(a)}^{g(b)} f(u) \mathrm{d}u\)