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education/math/MATH1210 (calc 1)/Integrals.md

@@ -104,4 +104,15 @@ $$ \dfrac{d}{dx} \int_{2x}^{3x} \sin(t) dt = \dfrac{d}{dx} -\cos(t)\Big|_{2x}^{3
 If $f(x)$ is continuous over an interval $[a, b]$ then there is at least one point $c$ in the interval $[a, b]$ such that:
 $$f(c) = \dfrac{1}{b-a}\int_a^bf(x)dx $$
 This formula can also be stated as $\int_a^b f(x)dx = f(c)(b-a)$
-This theorem tells us that a continuous function on the closed interval will obtain its average for at least one point in the interval.
+This theorem tells us that a continuous function on the closed interval will obtain its average for at least one point in the interval.
+
+# U-Substitution
+## Forumulas
+- $\int k {du} = ku + C$
+- $\int u^n du = \frac{1}{n+1}u^{n+1} + C$
+- $\int \frac{1}{u} du = \ln(|u|) + C$ 
+- $\int e^u du = e^u + C$
+- $\int \sin(u) du = -\cos(u) + C$
+- $\int \cos(u) du = \sin(u) + C$
+- $\int \dfrac{1}{\sqrt{a^2 - u^2}} du = \arcsin(\frac{u}{a}) +C$ 
+- $\int \dfrac{1}{a^2+u^2}du = \dfrac{1}{a} \arctan(\frac{u}{a}) + C$