Author Topic: Kanye East's ENG1091 (Uni Maths) thread (Read 1676 times) Tweet Share

Kanye East · « **on:** October 21, 2013, 03:16:32 pm »

I have another integration by parts question. I keep getting myself stuck in a loop when I do this question...
Integrate the following by using the Integration by parts method:
$\int\sin^2(x)dx$

This is what I did:
$\int\sin(x) sin(x)dx$
$=-sin(x)cos(x)-\int\((cos(x) *-cos(x))dx$
$=-sin(x)cos(x)+\int\cos^2(x)dx$
$Let A=\int\cos^2(x)dx$
$A=cos(x)sin(x)+\int\sin^2(x)dx <- WTF. Why am I stuck in a loop?$

b^3 · « **Reply #1 on:** October 21, 2013, 03:46:48 pm »

Kinda feels like overkill (there's an easier way without int by parts) but anyways, so that you don't get stuck in a loop, you need to make your $v'=1$ and $u=\sin^{2}(x)$ and then use int by parts twice.
$\begin{alignedat}{1}\int\sin^{2}\left(x\right)dx & =\int\sin^{2}\left(x\right)\frac{d}{dx}\left(x\right)dx \\ & =x\sin^{2}\left(x\right)-\int x\times2\sin\left(x\right)\cos\left(x\right)dx \\ & =x\sin^{2}\left(x\right)-\int x\sin\left(2x\right)dx \\ & =x\sin^{2}\left(x\right)-\int\left(\frac{d}{dx}\left(-\frac{1}{2}\cos\left(2x\right)\right)x\right)dx \\ & =x\sin^{2}\left(x\right)-\left(-\frac{1}{2}x\cos\left(2x\right)-\int-\frac{1}{2}\cos\left(2x\right)\cdot\left(1\right)dx\right) \\ & =x\sin^{2}\left(x\right)+\frac{1}{2}x\cos(2x)-\frac{1}{4}\sin\left(2x\right)+C,\: C\in\mathbb{R} \\ & =x\sin^{2}\left(x\right)+\frac{1}{2}x\left(1-2\sin^{2}\left(x\right)\right)-\frac{1}{2}\sin\left(x\right)\cos\left(x\right)+C,\: C\in\mathbb{R} \\ & =\frac{1}{2}\left(x-\sin\left(x\right)\cos\left(x\right)\right)+C,\: C\in\mathbb{R} \end{alignedat}$

the easier, non-int by parts method, if you're interested

$\begin{alignedat}{1}\cos\left(2x\right) & =1-2\sin^{2}\left(x\right) \\ \implies\sin^{2}\left(x\right) & =\frac{1-\cos\left(2x\right)}{2} \\ \int\sin^{2}\left(x\right)dx & =\int\frac{1-\cos\left(2x\right)}{2}dx \\ & =\int\frac{1}{2}dx-\int\frac{1}{2}\cos\left(2x\right)dx \\ & =\frac{1}{2}x-\frac{1}{4}\sin\left(2x\right)+C,\: C\in\mathbb{R} \\ & =\frac{1}{2}\left(x-\sin\left(x\right)\cos\left(x\right)\right)+C,\: C\in\mathbb{R} \end{alignedat}$

Kanye East · « **Reply #2 on:** October 21, 2013, 08:59:28 pm »

Use a substitution and an integration by parts to evaluate the following:
$\int\ (3x-7)sin(5x+2)dx$
Not sure what the relationship between $3x-7$ and $5x+2$ is in order to make a u substitution...

pi · « **Reply #3 on:** October 21, 2013, 09:10:02 pm »

Try a substitution after you've already expanded it out and begun to use int by parts

Kanye East · « **Reply #4 on:** October 26, 2013, 02:32:00 pm »

Phy124 · « **Reply #5 on:** October 26, 2013, 03:10:29 pm »

a) Let $A = (1, 2, -1)$ and $B = (2, 0, 3)$

$\vec{OP} = \vec{OA} \ + \ t \vec{AB} = (1, 2, -1) \ + \ t(2 - 1, 0 - 2, 3 + 1) = (1, 2, -1) \ + \ t(1, -2, 4) = (1 + t, 2 - 2t, -1 + 4t)$

$x = 1 + t$

$\Rightarrow t = 1-x$

$y = 2 - 2t$

$\Rightarrow t = \frac{2-y}{-2}$

$z = -1 + 4t$

$\Rightarrow t = \frac{-1 - z}{4}$

$\Rightarrow \frac{1-x}{1} = \frac{2-y}{-2} = \frac{-1 - z}{4}$

b) Assuming it's asking for the shortest distance that hint is quite odd, I would have thought it would be much more productive to use the cross product like so;

If we take $C = (1, 0, 1)$ and the direction vector of the line as $\vec{v} = (1, -2, 4)$ then we simply have;

$d = ||\vec{AC}||\sin\theta$

$= \frac{||\vec{AC}\times \vec{v}||}{||\vec{v}||}$

Then again, I haven't done any of this stuff in quite some time, so I could be way off

Kanye East · « **Reply #6 on:** October 26, 2013, 03:46:35 pm »

Thanks!

I have another question to do with parametric equations and planes =.="
Attached it to the photo below.
Edit: There's a line next to the 2, just ignore it. I think I clicked my mouse on that point when I used snipping tool lol

TrueTears · « **Reply #7 on:** October 26, 2013, 03:50:41 pm »

a) http://en.wikipedia.org/wiki/Line-plane_intersection
b) http://en.wikipedia.org/wiki/Plane_%28geometry%29#Defining_a_plane_through_three_points
c) http://en.wikipedia.org/wiki/Plane_%28geometry%29#Dihedral_angle

Kanye East · « **Reply #8 on:** November 02, 2013, 09:19:25 am »

Find the limit, if it exists, for the following:

TrueTears · « **Reply #9 on:** January 08, 2014, 10:40:14 pm »

Sorry for bringing up and old thread, but I think the solution to this is pretty useful and should be noted.

Define a sequence $(a_n)$ to be eventually in a set $A \subseteq \mathbb{R}$ if there exists an $N \in \mathbb{N}$ such that $a_n \in A$ for all $n \ge \mathbb{N}$ .

Then I claim $(a_n) = \exp(-n)$ eventually converges to $0$ . To see this, for any positive $\epsilon > 0$ , pick $N \in \mathbb{N}$ such that $N = -\ln(\epsilon)$ so that whenever $n \ge N$ , it follows $|a_n| < \epsilon$ .

So the finite terms don't really matter, eventually.

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