Author Topic: Recreational Problems (SM level) (Read 92739 times) Tweet Share

dcc · « **Reply #165 on:** May 25, 2008, 04:16:56 pm »

Another approach for #2:

$I = \int x \cdot \sqrt{x^4 + 1}\mbox{ }dx$

$\mbox{Let u = }x^2\mbox{, }\dfrac{du}{dx} = 2x$

$I = \int \dfrac{1}{2} \cdot \sqrt{u^2 + 1}\mbox{ }du$

$\mbox{Let }\tan v \mbox{ = }u\mbox{, }\dfrac{du}{dv} = \sec^2 v$

$I = \dfrac{1}{2} \int \sec^3 v\mbox{ }dv$

From above, we know the integral of sec^3[v]:

$I = \dfrac{1}{4} \sec v \tan v + \dfrac{1}{4}log_{e}|\sec v + \tan v| + c$

$I = \dfrac{1}{4} (\sqrt{u^2 + 1} \cdot u) + \dfrac{1}{4}log_{e}|\sqrt{u^2 + 1} + u| + c$

$\implies \int x \cdot \sqrt{x^4 + 1}\mbox{ }dx = \dfrac{1}{4} (\sqrt{x^4 + 1} \cdot x^2) + \dfrac{1}{4}log_{e}|\sqrt{x^4 + 1} + x^2| + c$

Mao · « **Reply #166 on:** May 25, 2008, 09:09:29 pm »

Quote from: dcc on May 25, 2008, 12:42:25 pm

I have fixed question 8, I had written the sign wrong.

good one! make me have to change everything =\

and the question is still stuffed! [i think]
[updated: http://vcenotes.com/forum/index.php/topic,11.msg40884.html#msg40884]

EDIT:
question 8 has been re-re-fixed, it now has a 4 in place, and it no longer is imaginary.
[updated - 2 : http://vcenotes.com/forum/index.php/topic,11.msg40884.html#msg40884]

Ahmad · « **Reply #167 on:** May 25, 2008, 09:53:54 pm »

Mao · « **Reply #168 on:** May 25, 2008, 10:12:29 pm »

Quote from: dcc on May 24, 2008, 12:23:32 pm

$\mbox{4. Find: } \int \dfrac{x \cdot \cos{2x} - \sin{2x}}{x^3}\mbox{ }dx$

we notice that

$\dfrac{x \cdot \cos{2x} - \sin{2x}}{x^3} = \frac{x^2 \cdot \cos{2x} - x\sin{2x}}{x^4}$

which dreadfully resembles the quotient rule:

hence, by intuition:

$\frac{d}{dx}\left( \frac{\sin{2x}}{x^2}\right) = 2\cdot \dfrac{x \cdot \cos{2x} - \sin{2x}}{x^3}$

$\implies \int \dfrac{x \cdot \cos{2x} - \sin{2x}}{x^3}\mbox{ }dx = \frac{\sin{2x}}{2x^2}+c$

Mao · « **Reply #169 on:** May 30, 2008, 08:23:09 pm »

Quote from: dcc on May 24, 2008, 12:23:32 pm

$\mbox{9. Find: } \int \dfrac{1}{x^4 + 1}\mbox{ }dx$

mmmmm

in search of an answer in a long and winded journey:

factorising $x^4+1=\left(x^2+\sqrt{2}x+1\right)\left(x^2-\sqrt{2}x+1\right)$

hence by partial fractions:

$\begin{align*} \frac{1}{x^4+1} &= \frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x+1\right} \\ 1 &= (Ax+B)\left(x^2-\sqrt{2}x+1\right) + (Cx+D)\left(x^2+\sqrt{2}x+1\right) \\ & \mbox{and after a bit of struggle....}\\ A &= \frac{\sqrt{2}}{4}\\ B &= \frac{1}{2}\\ C &= -\frac{\sqrt{2}}{4}\\ D &= \frac{1}{2}\\ \end{align*}$

$\begin{align*} \frac{1}{x^4+1} &= \frac{ \frac{ \sqrt{2} }{4} x+ \frac{1}{2} }{ x^2+\sqrt{2}x+1 } + \frac{ -\frac{ \sqrt{2} }{ 4 }x + \frac{ 1}{2} }{x^2-\sqrt{2}x+1} \\ &= \frac{\frac{\sqrt{2}}{4}x+\frac{1}{4}}{x^2+\sqrt{2}x+1} + \frac{\frac{1}{4}}{x^2+\sqrt{2}x+1} + \frac{-\left( \frac{\sqrt{2}}{4}x -\frac{1}{4} \right) }{x^2-\sqrt{2}x+1} + \frac{\frac{1}{4}}{x^2-\sqrt{2}x+1}\\ &= \frac{\frac{\sqrt{2}}{4}x+\frac{1}{4}}{x^2+\sqrt{2}x+1} + \frac{\frac{1}{4}}{\left(x+\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 } + \frac{-\left( \frac{\sqrt{2}}{4}x -\frac{1}{4} \right) }{x^2-\sqrt{2}x+1} + \frac{\frac{1}{4}}{\left(x-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 }\\ \end{align*}$

which I will refer to as $P(x),Q(x),R(x),S(x)$ respectively

$\int \frac{1}{x^4+1} dx = \int P(x)\ dx + \int Q(x)\ dx + \int R(x)\ dx + \int S(x)\ dx$

$\begin{align*} \int P(x)\ dx &= \int \frac{\frac{\sqrt{2}}{4}x+\frac{1}{4}}{x^2+\sqrt{2}x+1}\ dx \\ &= \frac{\sqrt{2}}{8} \int \frac{2x + \sqrt{2}}{x^2+\sqrt{2}x+1}\ dx\\ &= \frac{\sqrt{2}}{8} \int \frac{du}{u} \\ &= \frac{\sqrt{2}}{8} \cdot log_e |x^2+\sqrt{2}x+1| + c \end{align*}$

$\begin{align*} \int R(x)\ dx &= \int \frac{-\left( \frac{\sqrt{2}}{4}x -\frac{1}{4} \right) }{x^2-\sqrt{2}x+1}\ dx \\ &= -\frac{\sqrt{2}}{8} \int \frac{2x - \sqrt{2}}{x^2-\sqrt{2}x+1}\ dx\\ &= -\frac{\sqrt{2}}{8} \int \frac{du}{u} \\ &= -\frac{\sqrt{2}}{8} \cdot log_e |x^2-\sqrt{2}x+1| + c \end{align*}$

$\begin{align*} \int Q(x) dx &= \int \frac{\frac{1}{4}}{\left(x+\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 }\ dx \\ &= \frac{\sqrt{2}}{4}\int \frac{\frac{\sqrt{2}}{2}}{\left(x+\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 } dx \\ &= \frac{\sqrt{2}}{4} \cdot \tan ^{-1} \left(\frac{x+\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\right) +c\\ &= \frac{\sqrt{2}}{4} \cdot \tan ^{-1} \left(\sqrt{2}x + 1\right) +c \end{align*}$

$\begin{align*} \int S(x) dx &= \int \frac{\frac{1}{4}}{\left(x-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 }\ dx \\ &= \frac{\sqrt{2}}{4}\int \frac{\frac{\sqrt{2}}{2}}{\left(x-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 } dx \\ &= \frac{\sqrt{2}}{4} \cdot \tan ^{-1} \left(\frac{x-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\right) +c\\ &= \frac{\sqrt{2}}{4} \cdot \tan ^{-1} \left(\sqrt{2}x - 1\right) +c \end{align*}$

THEREFORE

$\int \frac{1}{x^4+1} dx = \frac{\sqrt{2}}{4} \cdot \left(\tan ^{-1}(\sqrt{2}x - 1)+ \tan ^{-1}(\sqrt{2}x + 1)\right)\ +\ \frac{\sqrt{2}}{8} \cdot \mbox{log_e} \left| \dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1} \right| + c$

I love you dcc

beezy4eva · « **Reply #170 on:** May 30, 2008, 09:02:49 pm »

dcc you really are evil in its purest form

dcc · « **Reply #171 on:** May 30, 2008, 10:18:52 pm »

excuse me, that is rude!

beezy4eva · « **Reply #172 on:** May 31, 2008, 09:20:51 pm »

Quote from: dcc on May 30, 2008, 10:18:52 pm

excuse me, that is rude!

but true

AppleXY · « **Reply #173 on:** June 01, 2008, 09:15:39 am »

Quote from: beezy4eva on May 31, 2008, 09:20:51 pm

Quote from: dcc on May 30, 2008, 10:18:52 pm
excuse me, that is rude!
but true

why?

droodles · « **Reply #174 on:** June 03, 2008, 06:38:14 pm »

i dont get it

Mao · « **Reply #175 on:** June 06, 2008, 06:38:09 pm »

reviving this awesome thread: a fairly easy problem out of my Calculus book,

If $f_0(x)=\frac{x}{x+1}$ and $f_{n+1}=f_{0}\circ f_{n}$ for $n\in \mathbb{N}_0$

Find an expression for $f_n$ .

- Solved by DivideBy0: http://vcenotes.com/forum/index.php/topic,11.msg42312.html#msg42312

/0 · « **Reply #176 on:** June 06, 2008, 07:37:38 pm »

Quote from: Mao on June 06, 2008, 06:38:09 pm

reviving this awesome thread: a fairly easy problem out of my Calculus book,

If $f_0(x)=\frac{x}{x+1}$ and $f_{n+1}=f_{0}\circ f_{n}$ for $n\in \mathbb{N}_0$

Find an expression for $f_n$ .

$f_1 = f_0 \circ f_0 = \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1}=\frac{x(x+1)}{(x+1)(2x+1)}=\frac{x}{2x+1}$

$f_2 = f_0 \circ f_0 \circ f_0= \frac{\frac{x}{x+1}}{2(\frac{x}{x+1})+1}=\frac{\frac{x}{x+1}}{\frac{3x+1}{x+1}} = \frac{x}{3x+1}$

Assume that $f_k = \frac{x}{(k+1)x+1}$ , $0 \leq k \leq n$

Consider $f_{n+1} = f_0 \circ f_n = \frac{\frac{x}{(n+1)x+1}}{\frac{x}{(n+1)x+1}+1}=\frac{\frac{x}{(n+1)x+1}}{\frac{x+(n+1)x+1}{(n+1)x+1}}=\frac{x}{((n+1)+1)x+1}$

Hence, by induction, $f_n = \frac{x}{(n+1)x+1}$ , $n \geq 0$

Nice easy problem that I can do, thanks Mao

dcc · « **Reply #177 on:** June 24, 2008, 08:33:04 pm »

Prove the Identity:

$\sin\left(ax\right) \cos\left(bx\right) = \dfrac{1}{2} \left(\sin \left(\left( a + b \right)x\right) + \sin \left(\left( a - b\right) x\right)\right)$

Mao · « **Reply #178 on:** June 24, 2008, 08:38:54 pm »

RHS: (assuming you dont want me to prove the compound angle formula~)

$\sin [(a+b)x ]= \sin (ax)\cos (bx) + \sin (bx) \cos(ax)$

$\sin[(a-b)x ]= \sin (ax)\cos(bx) - \sin(bx) \cos(ax)$

$RHS=\frac{1}{2}(\sin [(a+b)x ]+\sin [(a-b)x)]=\frac{1}{2}\cdot 2 \sin (ax)\cos (bx)=\sin (ax)\cos (bx)=LHS$

QED

ideally i should've started with the LHS, but its basically the same principle

Mao · « **Reply #179 on:** June 26, 2008, 08:03:44 pm »

$\mbox{The area enclosed by }y=x^2,\; x=1,\; x=2\mbox{ and the x axis is spun around the y axis}$

$\mbox{Find the volume of this solid.}$

$\mbox{The area enclosed by }y=x^2,\; y=x^3\; \mbox{ is spun around the y axis}$

$\mbox{Find the volume of this solid.}$

EDIT: second question updated. it was too terrible

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Author Topic: Recreational Problems (SM level) (Read 92739 times) Tweet Share

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