Cambridge Maths Academy

수학 모음 (Maths collection) 전체보기 For a function which depends on two variables $(x,y)$, $$ \textrm f = \textrm f(x,y) $$ we find the critical points by considering 2-dimensional gradient and second-order derivatives. (In 1D, the critical points are usually called the stationary points.) (i) Critical points: $$ \begin{align} \nabla \textrm f = \left( \frac{ \partial \textrm f }{ \partial x }, \frac{..

수학 모음 (Maths collection) 전체보기 This post has been partly motivated by P2 §11.6 Integration by parts CP2 §6.5 Integrating hyperbolic functions CP2 §7.1 First-order differential equations (obtaining the particular integrals using the integrating factor) Question. Derive the following results. $$ \begin{align} \textrm{(a)} &&& I_1(a,b) = \int \textrm e^{ ax } \cos b x \, \textrm dx = \frac{ \textrm ..

수학 모음 (Maths collection) 전체보기 This little investigation has been motivated by Exercise 7B, Challenge in CP2 §7.2 Second-order homogeneous differential equations. Question 1. What happens with a first-order differential equation with complex constant coefficients? $$ \begin{align} (a_1 + ia_2) \frac{ \textrm{d} y }{ \textrm{d}x } + (b_1 + ib_2) y &= 0 \end{align} $$ Method 1. Using matrix notatio..

수학 모음 (Maths collection) 전체보기 Question. Let $$ \begin{align} I_n = \int \sin^n(ax) \, \textrm{d}x \end{align} $$ (a) Show that $$ \begin{align} I_n = -\frac1{an} \sin^{n-1}(ax) \cos(ax) + \frac{n-1}{n}I_{n-2} \end{align} $$ Let $$ \begin{align} J_n = \int_0^{\frac{\pi}{2}} \sin^nx \, \textrm{d}x \end{align} $$ (b) Hence, or otherwise, show that $$ \begin{align} J_n = \frac{n-1}{n}J_{n-2} \end{al..

수학 모음 (Maths collection) 전체보기 Consider $$ \begin{align} J = \int \sqrt{ \cot x }\, \textrm{d}x \quad \textrm{for} \quad 0 < x \le \frac{\pi}{2} \end{align} $$ (a) Method 1. By using the substitution $u = \sqrt{ \cot x }$, followed by partial fractions, we can show that $$ \begin{align} J &= \frac1{2\sqrt{2}} \ln \left\vert \frac{1 + \sqrt{2 \cot x} + \cot x}{1 - \sqrt{2 \cot x} + \cot x} \right\..

수학 모음 (Maths collection) 전체보기 Consider $$ \begin{align} I = \int \sqrt{\tan x}\,\textrm{d}x \quad \textrm{for} \quad 0 \le x < \frac{\pi}{2} \end{align} $$ (a) Method 1. By using the substitution $u = \sqrt{ \tan x }$, followed by partial fractions, we can show that $$ \begin{align} I &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \tan x - \sqrt{2 \tan x} + 1 }{ \tan x + \sqrt{2 \tan x} + 1 } \right..

수학 모음 (Maths collection) 전체보기 1. $u={\rm csch}\,x+\coth x$ 치환적분 $$ \int{\rm csch}\,x\,\textrm{d}x=-\ln\vert{\rm csch}\,x+\coth x\vert+c $$ 첫 번째 시도로 코시컨트의 적분에서처럼 자연로그를 시도할 수 있다. 쌍곡시컨트(sech)와 달리 쌍곡코시컨트는 삼각함수와 쌍곡선함수의 도함수에 마이너스 차이가 없기 때문에 자연로그 적분이 가능하다는 것을 알게 된다. 우리에게 필요한 도함수들은 다음과 같다. $$ \begin{align} \frac{\textrm{d}}{\textrm{d}x}\csc x&=-\csc x\cot x & \frac{\textrm{d}}{\textrm{d}x}{\rm csch}\,x&..

수학 모음 (Maths collection) 전체보기 0. $\int{\rm sech}\,x\,\textrm{d}x=\ln\vert{\rm sech}\,x+\tanh x\vert+c\,$? 첫 번째 시도로 시컨트의 적분에서처럼 자연로그를 시도할 수 있다. 그러나 삼각함수와 쌍곡선함수의 도함수에서 마이너스 차이로 인해 가능하지 않다는 것을 알게 된다. 이 차이점을 아는 것도 중요하므로 이 부분부터 살펴보자. 우리에게 필요한 도함수들은 다음과 같다. $$ \begin{align} \frac{\textrm{d}}{\textrm{d}x}\tan x&=\sec^2x & \frac{\textrm{d}}{\textrm{d}x}\tanh x&={\rm sech}^2x \\ \frac{\textrm{d}}{\textrm..