- *Integrand* is $f$.
- $F(x)$ may be any integral, i.e. $c$ is inconsequential
+### Integration by substitution
+
+$$\int f(u) {du \over dx} \cdot dx = \int f(u) \cdot du$$
+
+Note $f(u)$ must be one-to-one $\implies$ one $x$ value for each $y$ value
+
+e.g. for $y=\int(2x+1)\sqrt{x+4} \cdot dx$:
+let $u=x+4$
+$\implies {du \over dx} = 1$
+$\implies x = u - 4$
+then $y=\int (2(u-4)+1)u^{1 \over 2} \cdot du$
+Solve as a normal integral
+
+#### Definite integrals by substitution
+
+For $\int^b_a f(x) {du \over dx} \cdot dx$, evaluate new $a$ and $b$ for $f(u) \cdot du$.
+
+### Trigonometric integration
+
+$$\sin^m x \cos^n x \cdot dx$$
+
+**$m$ is odd:**
+$m=2k+1$ where $k \in \mathbb{Z}$
+$\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x$
+Substitute $u=\cos x$
+
+**$n$ is odd:**
+$n=2k+1$ where $k \in \mathbb{Z}$
+$\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x$
+Subbstitute $u=\sin x$
+
+**$m$ and $n$ are even:**
+Use identities:
+
+- $\sin^2x={1 \over 2}(1-\cos 2x)$
+- $\cos^2x={1 \over 2}(1+\cos 2x)$
+- $\sin 2x = 2 \sin x \cos x
+
+
## Applications of antidifferentiation
- $x$-intercepts of $y=f(x)$ identify $x$-coordinates of stationary points on $y=F(x)$