Author Topic: Nagisa Maths Thread (Read 26733 times) Tweet Share

Nagisa · « **Reply #60 on:** December 27, 2012, 11:55:46 pm »

Quote from: Hancock on December 27, 2012, 11:05:06 pm

Just read that you wanted trig substitution. Try $x = tan(\theta)$

kk heres my solution with trig subs with $x = 3\tan \theta \ \ \ \ \ dx = 3\sec^2\theta \ d \theta$

$\int \frac{x^3}{\sqrt{x^2 + 9}} \ dx$

$\sqrt{9\tan^2\theta + 1} = \sqrt{9\sec^2\theta} = 3\sec\theta$

$\int \frac{27\tan^3\theta}{3\sec\theta} 3\sec^2 \ d\theta = \int 27\tan^3\theta . \sec \theta \ d\theta$

$27 \int tan^3\theta . \sec \theta \ d \theta = 27 \int tan^2\theta . \tan \theta \sec \theta \ d \theta = 27 \int (\sec^2\theta - 1) . \tan \theta \sec \theta \ d \theta$

now let $u = \sec \theta$ so that $du = \tan \theta \sec \theta \ d \theta$

$= 27 \int (u^2 - 1 ) \ du = 9u^3 - 27u + C= 9\sec^3\theta - 27\sec \theta + C$

so now with the triangle shit again which im not gonna draw we get $9 \left (\frac{\sqrt{x^2 + 9}}{3} \right )^{3} - \frac{27\sqrt{x^2+9}}{3} + C$

$= \frac{1}{3} (\sqrt{x^2 + 9})^3 - 9\sqrt{x^2 + 9} + C$

and this is same as hancock's, hope im correct

Nagisa · « **Reply #61 on:** December 28, 2012, 12:01:44 am »

Quote from: Indian Battman on December 27, 2012, 11:22:38 pm

Using a trig sub, see if this helps, although see if you can still get there on your own too
Spoiler
$\begin{alignedat}{1}\mathrm{Let\:}x=3\tan\theta \\ \frac{dx}{d\theta} & =3\sec^{2}\theta \\ x^{2}+9 & =9\left(\tan^{2}\theta+1\right) \\ & =9\sec^{2}\theta \\ \int\frac{x^{3}}{\sqrt{x^{2}+9}}dx & =\int\frac{27\tan^{3}\theta}{\sqrt{9\sec^{2}\theta}}3\sec^{2}\theta d\theta \\ & =\int\frac{27\tan^{3}\theta}{3\sec\theta}3\sec^{2}\theta d\theta \\ & =9\int\tan\theta\tan^{2}\theta\sec^{2}\theta d\theta \\ & =9\int\tan\theta\left(\sec^{2}\theta-1\right)\sec^{2}\theta d\theta \\ \mathrm{Let\:}u=\sec\theta \\ \frac{du}{d\theta} & =\tan\theta\sec\theta \\ & =9\int\left(u^{2}-1\right)udu \end{alignedat} $

You should be able to get it from there

EDIT: Also I forgot the domain restriction from the mod from the square root, but that should lead you on the right path anyway

e.g. $\sqrt{x^{2}+9}=\sqrt{9\sec^{2}(x)}=3|\sec(x)|=3\sec(x)$ for $x\in [-\frac{\pi}{2},\frac{\pi}{2}]$

looks like mine was a bit different to urs but still seems to work

b^3 · « **Reply #62 on:** December 28, 2012, 01:23:16 am »

Found my mistake, I did $\frac{\sec^{2}(\theta)}{\sec(\theta)}=\sec^{2}(\theta)$ instead of $\sec(\theta)$

Sigh...

Nagisa · « **Reply #63 on:** December 28, 2012, 01:26:30 am »

Quote from: Indian Battman on December 28, 2012, 01:23:16 am

Found my mistake, I did $\frac{\sec^{2}(\theta)}{\sec(\theta)}=\sec^{2}(\theta)$ instead of $\sec(\theta)$

Sigh...

haha happens to the best of us, i didnt see it either so hope u dont feel bad lol

JellyDonut · « **Reply #64 on:** December 28, 2012, 03:26:46 am »

for some reason the thought of nagisa doing maths is fucking hilarious

gj

Nagisa · « **Reply #65 on:** December 28, 2012, 03:40:35 am »

Quote from: JellyDonut on December 28, 2012, 03:26:46 am

for some reason the thought of nagisa doing maths is fucking hilarious

gj

haha thinking bout that is funny as hahaha,

Nagisa · « **Reply #66 on:** December 28, 2012, 07:51:16 pm »

Evaluate $\int \frac{x + 4}{x^2 + 2x + 5} \ dx$ with partial fractions

b^3 · « **Reply #67 on:** December 28, 2012, 07:58:46 pm »

Split it up
$\begin{alignedat}{1}\int\frac{x+4}{x^{2}+2x+5}dx & =\int\frac{x}{x^{2}+2x+5}dx+\int\frac{4}{x^{2}+2x+5}dx\end{alignedat}$
Now we can use a $u$ substitution on the first one, which you should be able to do, but for the second one, we need to turn it into something we can work with, so lets complete the square on the bottom.
$\begin{alignedat}{1}x^{2}+2x+5 & =x^{2}+2x+\left(1\right)^{2}-\left(1\right)^{2}+5 \\ & =\left(x+1\right)^{2}+4 \end{alignedat} $

So now we have
$\begin{alignedat}{1}\int\frac{4}{\left(x+1\right)^{2}+4}dx & =2\int\frac{2}{\left(x+1\right)^{2}+4}dx\end{alignedat}$
Which you should be able to integrate to get a $\tan^{-1}$ of some sort.

Hope that helps.

polar · « **Reply #68 on:** December 28, 2012, 08:04:51 pm »

Spoiler

$\begin{aligned} \int \frac{x+4}{x^2 + 2x + 5} dx &= \int \frac{x+4}{(x+1)^2+4}dx \\ \text{Let \;} u &= x+1, \; du=dx, \; x= u-1 \\ \int \frac{(u-1)+4}{u^2+4} du &= \int \frac{u+3}{u^2+4} du \\ &= \int \frac{u}{u^2+4} du + \int \frac{3}{u^2+4} du \\ &= \frac{1}{2}\int \frac{2u}{u^2+4} du + \frac{3}{2}\int \frac{2}{u^2+4} du \\ &= \frac{1}{2}\log_e{|u^2+4|} + \frac{3}{2}\arctan\left(\frac{u}{2}\right) + c \\ &= \frac{1}{2}\log_e{|(x+1)^2+4|} + \frac{3}{2}\arctan\left(\frac{x+1}{2}\right) + c \\ &= \frac{1}{2}\log_e{(x^2+2x+5)} + \frac{3}{2}\arctan\left(\frac{x+1}{2}\right) + c \; \; \; (\text{where \;} x^2 + 2x + 5 >0 \; \text{for \;} x \in R)\end{aligned}$

Nagisa · « **Reply #69 on:** December 28, 2012, 08:20:37 pm »

Quote from: Battman on December 28, 2012, 07:58:46 pm

Split it up
$\begin{alignedat}{1}\int\frac{x+4}{x^{2}+2x+5}dx & =\int\frac{x}{x^{2}+2x+5}dx+\int\frac{4}{x^{2}+2x+5}dx\end{alignedat}$
Now we can use a $u$ substitution on the first one, which you should be able to do, but for the second one, we need to turn it into something we can work with, so lets complete the square on the bottom.
$\begin{alignedat}{1}x^{2}+2x+5 & =x^{2}+2x+\left(1\right)^{2}-\left(1\right)^{2}+5 \\ & =\left(x+1\right)^{2}+4 \end{alignedat} $

So now we have
$\begin{alignedat}{1}\int\frac{4}{\left(x+1\right)^{2}+4}dx & =2\int\frac{2}{\left(x+1\right)^{2}+4}dx\end{alignedat}$
Which you should be able to integrate to get a $\tan^{-1}$ of some sort.

Hope that helps.

ty so much ur always a big help and not giving me the answer is awesome. keep it up

Nagisa · « **Reply #70 on:** December 28, 2012, 08:21:55 pm »

Quote from: polar on December 28, 2012, 08:04:51 pm

Spoiler
$\begin{aligned} \int \frac{x+4}{x^2 + 2x + 5} dx &= \int \frac{x+4}{(x+1)^2+4}dx \\ \text{Let \;} u &= x+1, \; du=dx, \; x= u-1 \\ \int \frac{(u-1)+4}{u^2+4} du &= \int \frac{u+3}{u^2+4} du \\ &= \int \frac{u}{u^2+4} du + \int \frac{3}{u^2+4} du \\ &= \frac{1}{2}\int \frac{2u}{u^2+4} du + \frac{3}{2}\int \frac{2}{u^2+4} du \\ &= \frac{1}{2}\log_e{|u^2+4|} + \frac{3}{2}\arctan\left(\frac{u}{2}\right) + c \\ &= \frac{1}{2}\log_e{|(x+1)^2+4|} + \frac{3}{2}\arctan\left(\frac{x+1}{2}\right) + c \\ &= \frac{1}{2}\log_e{(x^2+2x+5)} + \frac{3}{2}\arctan\left(\frac{x+1}{2}\right) + c \; \; \; (\text{where \;} x^2 + 2x + 5 >0 \; \text{for \;} x \in R)\end{aligned}$

tyvm man, always helpin me out ,giving me something to check when ive done my answer, everyone should use spoiler tags like u lol

BubbleWrapMan · « **Reply #71 on:** December 28, 2012, 08:29:49 pm »

Quote from: Nagisa on December 28, 2012, 08:20:37 pm

ty so much ur always a big help and not giving me the answer is awesome. keep it up

Quote from: Nagisa on December 28, 2012, 08:21:55 pm

tyvm man, always helpin me out ,giving me something to check when ive done my answer, everyone should use spoiler tags like u lol

You're starting to hurt my brain.

Nagisa · « **Reply #72 on:** December 28, 2012, 08:55:27 pm »

would u like a thankyou aswell climbtoohigh?

BubbleWrapMan · « **Reply #73 on:** December 28, 2012, 09:00:08 pm »

Sure why not

Lasercookie · « **Reply #74 on:** December 28, 2012, 09:04:35 pm »

Quote from: Calvin Climb on December 28, 2012, 08:29:49 pm

You're starting to hurt my brain.

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

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