From 657edfba3d101e7a98ccff533e695e3746f47852 Mon Sep 17 00:00:00 2001 From: arc Date: Sun, 13 Apr 2025 19:53:16 -0600 Subject: [PATCH] vault backup: 2025-04-13 19:53:16 --- education/math/MATH1210 (calc 1)/Integrals.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/education/math/MATH1210 (calc 1)/Integrals.md b/education/math/MATH1210 (calc 1)/Integrals.md index 04b10d6..1d2e206 100644 --- a/education/math/MATH1210 (calc 1)/Integrals.md +++ b/education/math/MATH1210 (calc 1)/Integrals.md @@ -107,7 +107,7 @@ 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. # U-Substitution -## Forumulas +## Formulas - $\int k {du} = ku + C$ - $\int u^n du = \frac{1}{n+1}u^{n+1} + C$ - $\int \frac{1}{u} du = \ln(|u|) + C$ @@ -115,4 +115,5 @@ This theorem tells us that a continuous function on the closed interval will obt - $\int \sin(u) du = -\cos(u) + C$ - $\int \cos(u) du = \sin(u) + C$ - $\int \dfrac{1}{\sqrt{a^2 - u^2}} du = \arcsin(\frac{u}{a}) +C$ -- $\int \dfrac{1}{a^2+u^2}du = \dfrac{1}{a} \arctan(\frac{u}{a}) + C$ \ No newline at end of file +- $\int \dfrac{1}{a^2+u^2}du = \dfrac{1}{a} \arctan(\frac{u}{a}) + C$ +- $\int \dfrac{1}{u\sqrt{u^2 - a^2}} du = \dfrac{1}{a}arcsec(\dfrac{|u|}{a}) + C$ \ No newline at end of file