Author Topic: TT's Maths Thread (Read 146116 times) Tweet Share

TrueTears · « **Reply #780 on:** January 19, 2010, 06:23:04 pm »

Quote from: kamil9876 on January 19, 2010, 05:02:12 pm

Quote from: TrueTears on January 17, 2010, 08:08:49 pm
Anyway a method I used to play around with this is:

let $f(k)=\sum{x=1}^k x^n$

$f(k+1)-f(k)=(k+1)^n$ This leads to the guess of a polynomial.

edit: oh yes, how could i forget. This can also be done combinatorily!

That's just telescoping

TrueTears · « **Reply #781 on:** January 20, 2010, 01:05:57 am »

kamil9876 · « **Reply #782 on:** January 20, 2010, 03:30:53 pm »

We want the volume of the shape that is the intersection of the sphere and cone, ie how much of the sphere enters the cone is the ammount of water spilt out.

This volume is generated by rotating the area of the intersection of circle and area around y axis:

$\pi \int_{c-r}^h f(y)^2 dy$

so really the only difficult part is finding the limits in terms of the variables given. We have that $c=\frac{r}{sin\theta}$ by geometry so the rest should be fine, just integrating.

TrueTears · « **Reply #783 on:** January 20, 2010, 09:39:11 pm »

Oh that's smart kamilz.

What about this one, just a bit stuck =X

Prove the reduction formula:

$\int \sec^n(x) dx = \frac{\tan(x)\sec^{n-2}(x)}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2}(x)dx \ \text{and} \ n \neq 1$

brightsky · « **Reply #784 on:** January 20, 2010, 10:10:22 pm »

"Split" the integral into:

$\int\sec^n (x) dx = \int \sec^2 x \sec^{n-2} x dx$

$\because \int \sec^2x dx = \tan(x) + c$

$\therefore \int \sec^n(x) dx = \int \sec^{n-2} x \frac{d}{dx} (\tan(x)) dx$

$= \tan(x)\sec^{n-2}(x) - \int \tan (x) \frac{d}{dx}(\sec^{n-2}(x) dx$

$= \tan (x) \sec^{n-2}(x) - \int \tan(x) [(n-2)\sec^{n-3} x (\sec(x) \tan(x))] dx$

$\because \tan^2x = \sec^2(x) - 1$

$\therefore \implies \tan(x) \sec^{n-2}(x) - (n-2) \int \sec^{n-2} x (\sec^2(x) -1) dx$

$= \tan(x) \sec^{n-2}(x) - (n-2) \int \sec^n(x) dx + (n-2) \int \sec ^{n-2}x dx$

Add $(n-2) \int \sec^n(x) dx$ to both sides so the LHS becomes $(n-2+1) \int \sec^n(x) dx = (n-1)\int \sec^n(x) dx$ .

Divide both sides by $(n-1)$ and you're done.

humph · « **Reply #785 on:** January 20, 2010, 10:24:03 pm »

$\sec^2 x = 1 + \tan^2 x$ , then make the substitution $u = \sec x$ .

TrueTears · « **Reply #786 on:** January 21, 2010, 12:04:12 am »

For trigonometric substitutions say we got :

$\sqrt{x^2-a^2}$ why do we make the substitution $x = a\sec(\theta) \ \forall \ 0 \le \theta < \frac{\pi}{2}$ or $\pi \le \theta < \frac{3\pi}{2}$

I know it "works" but what is the reason behind it?

I'm not after a proof, I'm just after a justification or an informal demonstration.

brightsky · « **Reply #787 on:** January 21, 2010, 12:21:41 am »

By the Pythagoras Theorem,

$\sqrt{x^2-a^2}$ suggests a triangle with a hypotenuse of $x$ and base length of $a$ .

This implies $\sec \theta = \frac{x}{a} \implies x = a \sec \theta$ .

$\sec \theta$ is positive in quadrants 4 and 1.

Hence why $0 \le \theta < \frac{\pi}{2}$ or $\pi \le \theta < \frac{3\pi}{2}$

TrueTears · « **Reply #788 on:** January 21, 2010, 09:58:19 am »

Yeah, it's pretty trivial to get sec, however that is not what I am after. You could have also done $y = \sqrt{x^2-a^2}$ so $y^2 +a^2 = x^2$ so $a^2 = x^2-y^2$ so $1 = \frac{x^2}{a^2} - \frac{y^2}{a^2}$ which is a hyperbola so the parametric equation $x = a\sec(\theta)$ would work. However that still doesn't explain why you make the substitution when you are integrating. Why do we do this kind of inverse substitution? I understand the restriction to make it 1-to-1 function or else the inverse would not be defined when changing the variable $\theta$ back into $x$ however the crux step is the first substitution, why does it work? (Not necessarily where it comes from)

What I'm looking for is analogous to say integration by parts which is the "reverse" of the product rule.

Oh wait nvm I got it, it's because by picking $x = a\sec(\theta)$ we can simplify the radical $x^2-a^2$ . It actually works quite similar to a parametric equation xD

TrueTears · « **Reply #789 on:** January 22, 2010, 10:13:03 pm »

Hmm I'm stuck at the end for this question...

$\int \tan^{-1}\left(\sqrt{x}\right) dx$

Let $u = \tan^{-1}\left(\sqrt{x}\right)$ and $dv = dx$

$\implies \int \tan^{-1}\left(\sqrt{x}\right) dx = x \tan^{-1}\left(\sqrt{x}\right) - \int \frac{x}{2\sqrt{x}\left(x+1\right)} dx$

So... how does one evaluate $\int \frac{x}{2\sqrt{x}\left(x+1\right)} dx$ ?

Damo17 · « **Reply #790 on:** January 22, 2010, 11:31:33 pm »

Quote from: TrueTears on January 22, 2010, 10:13:03 pm

Hmm I'm stuck at the end for this question...

$\int \tan^{-1}\left(\sqrt{x}\right) dx$

Let $u = \tan^{-1}\left(\sqrt{x}\right)$ and $dv = dx$

$\implies \int \tan^{-1}\left(\sqrt{x}\right) dx = x \tan^{-1}\left(\sqrt{x}\right) - \int \frac{x}{2\sqrt{x}\left(x+1\right)} dx$

So... how does one evaluate $\int \frac{x}{2\sqrt{x}\left(x+1\right)} dx$ ?

Well.. I recognised that $- \frac{x}{2\sqrt{x}(x+1)}= \frac{1}{2\sqrt{x}(x+1)} \ - \ \frac{1}{2\sqrt{x}}$ .

So take the integral of that and put it all together to get:

$\int \tan^{-1}\left(\sqrt{x}\right) dx= x \tan^{-1}\left(\sqrt{x}\right) + \tan^{-1}\left(\sqrt{x}\right)- \sqrt{x} \ + \ C =(x+1)\tan^{-1}\left(\sqrt{x}\right)- \sqrt{x} \ + \ C$

TrueTears · « **Reply #791 on:** January 22, 2010, 11:39:36 pm »

Thanks Damo! That was awesome!

Yeah u substitution could have also worked by letting $u = \sqrt{x}$

TrueTears · « **Reply #792 on:** January 25, 2010, 08:17:54 pm »

$\int \frac{1}{1-e^{2x}}dx$

hmmmz

brightsky · « **Reply #793 on:** January 25, 2010, 08:29:23 pm »

Quote from: TrueTears on January 25, 2010, 08:17:54 pm

$\int \frac{1}{1-e^{2x}}dx$

hmmmz

Substitute $u = 2x, du = 2dx$ :

$\frac{1}{2} \int \frac{1}{1-e^u} du$

Substitute $v = e^u, dv = e^u du$

$\frac{1}{2} \int \frac{1}{v} \cdot \frac{1}{1-v} dv$

$\frac{1}{2} \int \left(\frac{1}{v} - \frac{1}{v-1}\right) dv$

Substitute $w = s - 1, dw = 1 dv$

$\frac{1}{2} \int \left(\frac{1}{v} - \frac{1}{w} \right) dw$

Integrate then substitute back in.

TrueTears · « **Reply #794 on:** January 25, 2010, 08:38:44 pm »

Yeap, good method, I just did $u = e^x$ and then partial fractions.

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