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 $
數學 Core
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M2 (代數&微積分)
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MC 技巧特訓
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傳授必學砌數技巧
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A1 A2 概念重構
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深入教授微積分技術
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拆解 Section B killer 題
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fx3650P 輕鬆屈機
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培訓考試技巧，重建答題框架
一對一/小組上課 收費詳情請 pm 查詢
(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 $
