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diff --git a/spec/calculus.md b/spec/calculus.md
index 9025d374ff7d109681790eed2e2c04cc07df36b2..b89eae68a78eddb15321a703067953204a1ade78 100644 (file)
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -254,6 +254,61 @@ $$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)$$
 - *Integrand* is $f$.
 - $F(x)$ may be any integral, i.e. $c$ is inconsequential
 
+#### Properties
+
+$$\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx$$
+
+$$\int^a_a f(x) \> dx = 0$$
+
+$$\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx$$
+
+$$\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx$$
+
+$$\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx$$
+
+### 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:
+
+- $\sin^2x={1 \over 2}(1-\cos 2x)$
+- $\cos^2x={1 \over 2}(1+\cos 2x)$
+- $\sin 2x = 2 \sin x \cos x$
+
+## Partial fractions
+
+On CAS: Action $\rightarrow$ Transformation $\rightarrow$ `expand/combine`
+
+
 ## Applications of antidifferentiation
 
 - $x$-intercepts of $y=f(x)$ identify $x$-coordinates of stationary points on $y=F(x)$
@@ -283,3 +338,11 @@ $$f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functi
 - when two graphs have the same ordinate, $y$-coordinate is double the ordinate
 - when two graphs have opposite ordinates, $y$-coordinate is 0 i.e. ($x$-intercept)
 - when one of the ordinates is 0, the resulting ordinate is equal to the other ordinate
+
+## Fundamental theorem of calculus
+
+If $f$ is continuous on $[a, b]$, then
+
+$$\int^b_a f(x) \> dx = F(b) - F(a)$$
+
+where $F$ is any antiderivative of $f$