Theorem:
\[\int f(g(x))g'(x)dx = \int fudu\] where \(u = g(x)\)
Proof:
Let \(F(u)\) be an antiderivative for \(f(u)\) and \(u = g(x)\) then \(\frac{dF(u)}{du} = f(u)\), and \(\frac{dF(u)}{dx} = \frac{dF(u)}{du} \cdot \frac{du}{dx} = f(g(x))g'(x)\) \(\Rightarrow \int \frac{dF(u)}{dx} dx = \int f(g(x))g'(x) dx\)
Example 1. \(\int (x^2-1)^42xdx\)
Let \(u = x^2-1\), then \(\frac{du}{dx} = 2x\),
then \(\int (x^2-1)^42xdx = \int u^4 \frac{du}{dx}dx = \int u^4du = \frac{1}{5}u^5 = \frac{1}{5}(x^2-1)^5+C\)
Example 2. \(\int sin(x)cos(x)dx\)
Let \(u = sin(x)\), then \(\frac{du}{dx} = cos(x)\),
then \(\int u\frac{du}{dx} = \int u \ du = \frac{1}{2}u^2+C = \frac{1}{2}sin^2(x)+C\)
Example 3. \(\int \frac{1}{(3+5x)^2}dx\)
Let \(u = 3+5x\), then \(\frac{du}{dx} = 5\),
then \(\int \frac{1}{(3+5x)^2}dx = \frac{1}{5} \int \frac{1}{u^2}\frac{du}{dx}dx = \frac{1}{5}\cdot (-1) u^{-1}+C=\frac{-1}{5}\cdot \frac{1}{3+5x}+C\)
Theorem:
\[\int_a^b f(x)g'(x)dx = f(x)g(x)\ |_a^b-\int_a^b f'(x)g(x)dx\]
Proof:
This is derived from the rule of multiplication of taking derivatives
\(\because[f(x)g(x)]' = f'(x)g(x)+f(x)g'(x)\) \(\therefore \int f(x)g'(x)dx = f(x)g(x)-\int f'(x)g(x)dx\)
Example 1. \(\int xe^x\ dx\)
Let \(f(x) = x\) and \(g'(x) = e^x\)
then \(\int xe^x\ dx = xe^x - \int 1 \cdot e^x\ dx = xe^x - e^x +C\)
Example 2. \(\int x^3ln(x)\ dx\)
Let \(f(x) = x^3\) and \(g'(x) = ln(x)\)
then \(\int x^3ln(x)\ dx = \int ln(x)x^3\ dx = ln(x)\cdot\frac{1}{4}x^4 - \int \frac{1}{x} \frac{1}{4}x^4\ dx = \frac{1}{4}ln(x)x^4 - \frac{1}{20}x^5 + C\)
Integrals involving \(\sqrt{a^2+x^2}\) or \(\sqrt{x^2-a^2}\), use trigonometric substitution
Example 1. \(\int \frac{1}{(9-x^2)^{3/2}}\ dx\)
Let \(x = 3sin(u)\), then \(\frac{dx}{du} = 3cos(u)\) and
\(\int \frac{1}{(9-x^2)^{3/2}}\ dx = \int \frac{1}{(9sin^2(u))^{3/2}}\cdot 3cos(u)\ du = \frac{1}{9}tan(u)+C = \frac{1}{9}\frac{x}{\sqrt{9-x^2}}+C\)