Author Topic: TT's Maths Thread (Read 146016 times) Tweet Share

Ahmad · « **Reply #795 on:** January 25, 2010, 09:16:51 pm »

Alternative approach re-using this idea.

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

and $J = \int \frac{e^x}{e^{-x} - e^{x}} dx$

$I + J = -\log (e^{-x}-e^x)$

$I - J = x$

$I = \frac{1}{2} (x - \log (e^{-x}-e^x))$

(upto constants)

brightsky · « **Reply #796 on:** January 25, 2010, 09:24:40 pm »

Quote from: Ahmad on January 25, 2010, 09:16:51 pm

Alternative approach re-using this idea.

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

and $J = \int \frac{e^x}{e^{-x} - e^{x}} dx$

$I + J = -\log (e^{-x}-e^x)$

$I - J = x$

$I = \frac{1}{2} (x - \log (e^{-x}-e^x))$

(upto constants)

Oohh that is smart...

TrueTears · « **Reply #797 on:** January 26, 2010, 12:23:22 am »

$\int \frac{1+\sin(x)}{1-\sin(x)} dx$

I've played around for a while, tried to multiply by $\frac{1 \pm \sin(x)}{1 \pm \sin(x)}$ ...

humph · « **Reply #798 on:** January 26, 2010, 07:10:33 am »

Haven't had a stab at it, but think about double/half-angle formulae, or try making the classic $t=tan(x/2)$ substitution (which works with most integrals of rational functions of trigonometric functions).

TrueTears · « **Reply #799 on:** January 26, 2010, 09:19:36 am »

Oh yeah I got it, thanks

Ahmad · « **Reply #800 on:** January 26, 2010, 01:12:06 pm »

My first intuition was to either multiply and divide by $1-\sin(x)$ or $1+\sin(x)$ , with the idea being to exploit $\sin^2(x) + \cos^2(x) = 1$ . I wrote down both expressions then discarded the first because it wasn't clear how to proceed, the second gives:

$\int \frac{(1+\sin(x))^2}{\cos^2(x)} dx$

$\int sec^2(x) + 2\underbrace{\frac{\sin(x)}{\cos^2(x)}}_{\frac{d}{dx} \frac{1}{\cos(x)}} + \underbrace{\tan^2{x}}_{\sec^2(x) - 1} dx$

$= 2 \tan(x) + 2 \sec x - x + C$ .

In the general case of rational functions of trigonometric functions as humph mentioned the idea of the t substitution is the important one (basically converts it into a rational function), downside being it can get messy fast.

TrueTears · « **Reply #801 on:** January 26, 2010, 01:19:24 pm »

$\int \frac{\sqrt{1+x^2}}{x^2} dx$

I did:

Let $x = \tan(\theta)$ where $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$

$\sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)$

$dx = \sec^2(\theta) d\theta$

$\int \frac{\sec(\theta)}{\tan^2(\theta)}\sec^2(\theta) d\theta = \int \frac{\sec^3(\theta)}{\tan^2(\theta)} d\theta$

Hmm now what?

brightsky · « **Reply #802 on:** January 26, 2010, 01:33:44 pm »

$\int \frac{\sec(\theta)}{\tan^2\(\theta)} \sec^2(\theta) d\theta = \int \frac{\sec^2(\theta)}{\tan^2\(\theta)} \sec(\theta) d\theta= \int \csc^2(\theta) \sec(\theta) d\theta = \int (\cot^2(\theta) + 1) \sec(\theta) d\theta$ etc....

TrueTears · « **Reply #803 on:** January 26, 2010, 10:21:09 pm »

$\int \frac{\sqrt{x}}{1+x^3} dx$

Let $u = \sqrt{x} \implies 2udu = dx$

$u^2 = x \implies u^6+1 = x^3+1$

$\therefore 2 \int \frac{u^2}{u^6+1} du$

I don't think this is the right way... to partial fraction that would take ages...

Damo17 · « **Reply #804 on:** January 26, 2010, 10:48:28 pm »

Quote from: TrueTears on January 26, 2010, 10:21:09 pm

$\int \frac{\sqrt{x}}{1+x^3} dx$

Let $u = \sqrt{x} \implies 2udu = dx$

$u^2 = x \implies u^6+1 = x^3+1$

$\therefore 2 \int \frac{u^2}{u^6+1} du$

I don't think this is the right way... to partial fraction that would take ages...

Let $u= \sqrt{x^3}$ , a far easier way.

Over9000 · « **Reply #805 on:** January 27, 2010, 03:06:31 pm »

THA FUKN PUSSAYY

$\int_0^3 \frac{dx}{x-1}$

TrueTears · « **Reply #806 on:** January 27, 2010, 10:45:28 pm »

Find the arc length function for the curve $y = x^2 - \frac{1}{8}\log_e(x)$ taking $(1,1)$ as the starting point.

So $s(x) = \int_1^x \sqrt{1+\left[f'(t)\right]^2} dt$

Let $f(t) = t^2 - \frac{1}{8}\log_e(t)$

$1+\left[f'(t)\right]^2 = \left(2t + \frac{1}{8t}\right)^2$

$\implies \sqrt{1+\left[f'(t)\right]^2} = \sqrt{\left(2t + \frac{1}{8t}\right)^2} = \left | 2t + \frac{1}{8t} \right |$

So then we need $s(x) = \int_1^x \left | 2t + \frac{1}{8t} \right | dt$

But my book only has: $s(x) = \int_1^x 2t + \frac{1}{8t} dt$ why did they not include the modulus signs?

/0 · « **Reply #807 on:** January 27, 2010, 11:05:42 pm »

Because in $[1,x]$ , $2t+\frac{1}{8t}$ is always positive.

TrueTears · « **Reply #808 on:** January 27, 2010, 11:06:45 pm »

But don't you need to state it though?

humph · « **Reply #809 on:** January 28, 2010, 08:49:04 am »

Why would you bother? It's a matter of simplification.

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