Everyday Math 每日數學 之 Integration #20190529 (+solution)

Let $u=\sqrt{x}$
Then $x=u^2$
$dx=2udu$
$\int \sec^2{\sqrt{x}}dx$
$=\int 2u\sec^2 udu$
$=\int 2ud(\tan u)$
$=2u\tan u-2\int \tan u du$
$=2u\tan u-2\int \frac{\sin u}{\cos u} du$
$=2u\tan u+2\int \frac{1}{\cos u} d(\cos u)$
$=2u\tan u+2\ln \left|\cos u \right| +C$
$=2\sqrt{x}\tan \sqrt{x}+2\ln \left|\cos \sqrt{x} \right| +C$

Everyday Math 每日數學 之 Statistics (Change of Data) #20190522 (+Solution)

Everyday Math 每日數學 之 Mathematical Induction + Trigonometry + Differentiation #20190601 (+Solution)

Solution:
(a)
For any non-negative integer $n$ , denoted by $P(n)$ the proposition that $\cos x \cos (2x) \cos(4x) ... \cos(2^n x)=\frac{\sin(2^{n+1}x)}{2^{n+1}\sin x}$ .
Note that $\cos x=\frac{2\sin x \cos x}{2\sin x}=\frac{\sin(2x)}{2\sin x}$ .
$P(0)$ is true.
Let $k$ be a non-negative integer. Assume $P(k)$ is true.
Therefore, $\cos x \cos (2x) \cos(4x) ... \cos(2^k x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin x}$ .
Note that $\cos x \cos (2x) \cos(4x) ... \cos(2^k x)\cos(2^{k+1}x)=\frac{\sin(2^{k+1}x)\cos(2^{k+1}x)}{2^{k+1}\sin x}=\frac{\sin(2^{k+2}x)}{2^{k+2}\sin x}$
Therefore, $P(k+1)$ is true.
By the principle of mathematical induction, for any non-negative integer $n$ , $\cos x \cos (2x) \cos(4x) ... \cos(2^n x)=\frac{\sin(2^{n+1}x)}{2^{n+1}\sin x}$

(b) $\frac{d}{dx} \left[\frac{1}{2^{101}}\sin(2^{101}x)\csc x \right]$
$=-\frac{1}{2^{101}}\sin(2^{101}x)\csc x\cot x+\frac{1}{2^{101}}\csc x \cos(2^{101}x)\cdot 2^{101}$
$=-\frac{1}{2^{101}}\sin(2^{101}x)\csc x\cot x+\csc x \cos(2^{101}x)$
$\frac{d}{dx}|_{x=\frac{\pi}{4}} \frac{1}{2^{101}}\sin(2^{101}x)\csc x$
$=-\frac{1}{2^{101}}\sin(2^{101}\cdot \frac{\pi}{4})\csc \frac{\pi}{4}\cot \frac{\pi}{4}+\csc \frac{\pi}{4} \cos(2^{101}\cdot \frac{\pi}{4})$
$=\sqrt 2$

全新 Fergus Sir DSE 2020 / DSE 2021 數學 CORE / M2 備戰課程

數學 Core	M2 (代數&微積分)
MC 技巧特訓	傳授必學砌數技巧
A1 A2 概念重構	深入教授微積分技術
拆解 Section B killer 題	fx3650P 輕鬆屈機

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Everyday Math 每日數學 之 Quadratic Equation and Graph 二次方程及圖像 #20190515 (+Solution)

Answer : D

Everyday Math 每日數學 之 Integration #20190517 (+Solution)

(a)
$e^x=\frac{1}{t-1}$
$x=-\ln(t-1)$
$dx=-\frac{1}{t-1}dt$
$\int \frac{1}{e^{4x}+e^{3x}}dx$
$=\int \frac{1}{\left(\frac{1}{t-1}\right)^{4}+\left(\frac{1}{t-1}\right)^{3}}\cdot -\frac{1}{t-1}dt$
$=-\int \frac{1}{\left(\frac{1}{t-1}\right)^{3}+\left(\frac{1}{t-1}\right)^{2}} dt$
$=-\int \frac{(t-1)^3}{1+t-1} dt$
$=-\int \frac{t^3-3t^2+3t-1}{t} dt$
$=-\frac{1}{3}t^3+\frac32t^2-3t+\ln|t|+C$
$=-\frac{1}{3}\left( \frac{1}{e^x}+1\right)^3+\frac32\left( \frac{1}{e^x}+1\right)^2-3\left( \frac{1}{e^x}+1\right)+\ln\left| \frac{1}{e^x}+1\right|+C$

(b)
Let $u=e^x$ . Then $du=e^x dx$ .
$\int \frac{1}{u^5+u^4}du$
$=\int \frac{1}{e^{5x}+e^{4x}}\cdot e^xdx$
$=\int \frac{1}{e^{4x}+e^{3x}}dx$
$=-\frac{1}{3}\left( \frac{1}{e^x}+1\right)^3+\frac32\left( \frac{1}{e^x}+1\right)^2-3\left( \frac{1}{e^x}+1\right)+\ln\left| \frac{1}{e^x}+1\right|+C$ ( by (a) )
$=-\frac{1}{3}\left( \frac{1}{u}+1\right)^3+\frac32\left( \frac{1}{u}+1\right)^2-3\left( \frac{1}{u}+1\right)+\ln\left|\frac{1}{u}+1\right|+C$

(c)
$\int _1^e \frac{5u^5+4u^4}{(u^5+u^4)^2}du$
$=\int _1^e \frac{(5u^4+4u^3)u}{(u^5+u^4)^2}du$
$=\int _{u=1}^{u=e} \frac{u}{(u^5+u^4)^2}d(u^5+u^4)$
$=\int _{u=1}^{u=e} u(u^5+u^4)^{-2}d(u^5+u^4)$
$=-\int _{u=1}^{u=e} ud(u^5+u^4)^{-1}$
$=- \left[ u(u^5+u^4)^{-1}\right]_{u=1}^{u=e} +\int _{u=1}^{u=e} (u^5+u^4)^{-1}du$
$=- \left[ (u^4+u^3)^{-1}\right]_{u=1}^{u=e} +\left[-\frac{1}{3}\left( \frac{1}{u}+1\right)^3+\frac32\left( \frac{1}{u}+1\right)^2-3\left( \frac{1}{u}+1\right)+\ln\left| \frac{1}{u}+1\right|\right]_{u=1}^{u=e}$
$=- (e^4+e^3)^{-1} -\frac{1}{3}\left( \frac{1}{e}+1\right)^3+\frac32\left( \frac{1}{e}+1\right)^2-3\left( \frac{1}{e}+1\right)+\ln\left( \frac{1}{e}+1\right)+\frac{19}{6}-\ln 2$

M2 Review - Chapter 13 Vector