Author Topic: Nagisa Maths Thread (Read 21355 times) Tweet Share

Nagisa · « **Reply #75 on:** December 28, 2012, 09:11:21 pm »

Quote from: laserblued on December 28, 2012, 09:04:35 pm

I think Nagisa liked how b^3 just pointed him in the right direction and then polar followed up by providing the full working in spoiler tags.

Anyway, try stay on-topic etc. this thread has been going pretty well for the most part

yes ever since i joined AN polar has been there, and b^3 has helped me ever since i started this thread just wanted to make sure they knew i was grateful. ty laseredd ur also nice lol

Nagisa · « **Reply #76 on:** December 29, 2012, 12:06:31 am »

Evaluate $\int \sqrt{\frac{1 + x}{1 - x}} \ dx$

brightsky · « **Reply #77 on:** December 29, 2012, 12:07:52 am »

hint: multiply top and bottom by sqrt(1+x) and then split.

Nagisa · « **Reply #78 on:** December 29, 2012, 12:21:57 am »

Quote from: brightsky on December 29, 2012, 12:07:52 am

hint: multiply top and bottom by sqrt(1+x) and then split.

ty

Nagisa · « **Reply #79 on:** December 29, 2012, 01:00:01 am »

okay thankyou brightsky

heres how i did it

$\int \sqrt \frac{1+x}{1-x}} \ dx$

$\int \sqrt \frac{1+x}{1-x}} . \frac{\sqrt{1+x}}{\sqrt{1+x}}\ dx$

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

so it becomes really simple lol ty brightsky. we do a simple sub on the second part $u=1-x^2 \ \ \ \ \ so \ du=-2x \ dx$

$=\sin^{-1}x - \sqrt{1-x^2} + C$

Planck's constant · « **Reply #80 on:** December 29, 2012, 01:17:51 am »

Quote from: brightsky on December 29, 2012, 12:07:52 am

hint: multiply top and bottom by sqrt(1+x) and then split.

Nice trick.
Another is the substitution x=cos2a, the integrand quickly simplifies to cota, and dx/da = -4sina*cosa etc

Nagisa · « **Reply #81 on:** December 29, 2012, 02:52:06 am »

Evaluate $\int_{-1}^{1} \frac{e^{\tan^{-1}x}}{1+x^2} \ dx$

TrueTears · « **Reply #82 on:** December 29, 2012, 03:13:44 am »

Spoiler

$u = \tan^{-1}(x)$

Nagisa · « **Reply #83 on:** December 29, 2012, 03:21:45 am »

okay TT wat bout this one $\int \frac{1}{x\sqrt{4x + 1}} \ dx$

Phy124 · « **Reply #84 on:** December 29, 2012, 04:12:34 am »

Try a double substitution, firstly let $u = 4x+1$ to produce a new integral, then secondly let $a = \sqrt{u}$ (or whatever letter you want to use for your second substitution) to simplify it again.

If you have any trouble with that I can write a full solution for you

Planck's constant · « **Reply #85 on:** December 29, 2012, 11:11:51 am »

Quote from: Clifford on December 29, 2012, 04:12:34 am

Try a double substitution, firstly let $u = 4x+1$ to produce a new integral, then secondly let $a = \sqrt{u}$ (or whatever letter you want to use for your second substitution) to simplify it again.

If you have any trouble with that I can write a full solution for you

Yes. In effect it is the single substitution u= sqrt(4x+1).
Everything goes nicely, and the answer pops out in a single step, as long as you can recognise the integral as that of arctanh(u)

polar · « **Reply #86 on:** December 29, 2012, 11:42:50 am »

here you go. this solution doesn't use the arctanh function though, so it's a bit longer

Spoiler

$\begin{aligned} \int \frac{1}{x\sqrt{4x+1}}\; dx \\ \text{Let \;} u&=\sqrt{4x+1}, \; x= \frac{u^2-1}{4} \\ \frac{du}{dx} &= \frac{2}{\sqrt{4x+1}} \\ &= \frac{2}{u} \\ dx &= \frac{u}{2}du \\ \; \\ \int \frac{1}{x\sqrt{4x+1}} dx &= \int \frac{1}{\frac{u^2-1}{4}\times{u}}\times\frac{u}{2}du \\ &= \int \frac{4}{u^2-1} \times \frac{1}{2} du \\ &= \int \frac{2}{u^2-1} du \\ &= \int \frac{1}{u-1} - \frac{1}{u+1} du \\ &= \log_e{|u-1|} - \log_e{|u+1|} + c \\ &= \log_e{\left|\frac{u-1}{u+1}\right|} + c \\ &= \log_e{\left|\frac{\sqrt{4x+1}-1}{\sqrt{4x+1}+1}\right|} + c\end{aligned}$

Planck's constant · « **Reply #87 on:** December 29, 2012, 12:17:47 pm »

Quote from: polar on December 29, 2012, 11:42:50 am

here you go. this solution doesn't use the arctanh function though, so it's a bit longer

Spoiler
$\begin{aligned} \int \frac{1}{x\sqrt{4x+1}}\; dx \\ \text{Let \;} u&=\sqrt{4x+1}, \; x= \frac{u^2-1}{4} \\ \frac{du}{dx} &= \frac{2}{\sqrt{4x+1}} \\ &= \frac{2}{u} \\ dx &= \frac{u}{2}du \\ \; \\ \int \frac{1}{x\sqrt{4x+1}} dx &= \int \frac{1}{\frac{u^2-1}{4}\times{u}}\times\frac{u}{2}du \\ &= \int \frac{4}{u^2-1} \times \frac{1}{2} du \\ &= \int \frac{2}{u^2-1} du \\ &= \int \frac{1}{u-1} - \frac{1}{u+1} du \\ &= \log_e{|u-1|} - \log_e{|u+1|} + c \\ &= \log_e{\left|\frac{u-1}{u+1}\right|} + c \\ &= \log_e{\left|\frac{\sqrt{4x+1}-1}{\sqrt{4x+1}+1}\right|} + c\end{aligned}$

Yes, that's right.
But you can stop 5 steps earlier if you recognise arctanh, else you end up deriving a formula for arctanh from first principles

Phy124 · « **Reply #88 on:** December 29, 2012, 05:53:59 pm »

Quote from: argonaut on December 29, 2012, 11:11:51 am

Yes. In effect it is the single substitution u= sqrt(4x+1).
Everything goes nicely, and the answer pops out in a single step, as long as you can recognise the integral as that of arctanh(u)

Yeah that's quite right, I just found teaching these sort of substitutions in multiple steps reduced errors made by the students, do whatever you feel most comfortable with

Nagisa · « **Reply #89 on:** December 29, 2012, 06:56:12 pm »

thanks guys, good to see u making an effort argonaut.

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