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