From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Thu, 25 Apr 2019 04:49:26 +0000 (+1000)
Subject: [spec] integration by substitution
X-Git-Tag: yr12~158
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/f05acf18e1c0d59f1fe02c6254e4eac4b9271084

[spec] integration by substitution
---

diff --git a/spec/calculus.md b/spec/calculus.md
index 9025d37..0121899 100644
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -254,6 +254,45 @@ $$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)$$
 - *Integrand* is $f$.
 - $F(x)$ may be any integral, i.e. $c$ is inconsequential
 
+### 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 $y=F(x)$