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$
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 $
