Author Topic: Nagisa Maths Thread (Read 21353 times) Tweet Share

Nagisa · « **Reply #45 on:** December 27, 2012, 12:44:59 am »

so my new solution is..

Evaluate $\int \tan^5x \ dx$

$\int \tan^5x \ dx = \int \tan x \tan^2x \tan^2x \ dx = \int \tan x \ (\sec^2 - 1) (\sec^2 - 1) \ dx$

$\int \tan x \ (\sec^4x - 2\sec^2x + 1) \ dx$

$\int \tan x \sec^4x - \tan x \ 2\sec^2x + \tan x \ dx$

$\int (\tan x \sec x)(\sec^3x) - (\tan x \sec x)(2\sec x) + \tan x \ dx$

$\int (\tan x \sec x)(\sec^3x) - (\tan x \sec x)(2\sec x) \ dx + \int \tan x \ dx$

$let \ u = \sec x \ \ \ \ \ \ \ so \ du = \sec x \tan x \ dx$

$\int u^3 \ du - \int 2u \ du + \int \tan x \ dx$ <--- thnx to b^3

$\left [\frac{u^4}{4} - u^2 + \ln |\sec x | + C\right ]$

$\frac{1}{4}\sec^4x - \sec^2x + \ln |\sec x | + C$

okay now is this okay??

BubbleWrapMan · « **Reply #46 on:** December 27, 2012, 01:00:20 am »

Looks good, the square brackets on the second last line aren't needed though, and you should use more '=' signs.

Nagisa · « **Reply #47 on:** December 27, 2012, 09:41:54 pm »

Evaluate the integral $\int \frac{1}{x^2\sqrt{x^2 - 9}} \ dx$

$let \ x = 3\sec \theta \ \ \ \ \ \ so \ dx = 3\sec \theta \tan \theta \ d\theta$

now apply back to the question.. specifically the $\sqrt{x^2 - 9}$

$\sqrt{x^2 - 9}$ = $\sqrt{9\sec^2\theta - 9}$ = $\sqrt{9(\sec^2\theta - 1)$ = $\sqrt{9\tan^2\theta}$ = $3 |\tan \theta| = 3\tan \theta$

$\therefore \int \frac{3\sec \theta \tan \theta}{9\sec^2\theta . 3\tan \theta} \ d \theta$ = $\frac{1}{9} \int \frac{1}{\sec \theta} \ d \theta$ = $\frac{1}{9} \sin \theta + C$

but this in terms of $\theta$ and the question is in terms of $x$ ... So remember when when we let $x = 3\sec \theta$ if we rearrange it we get $\sec \theta = \frac{x}{3}$ now this means we can make a magical triangle.. see below.

so now we can apply our integral that was in terms of $\theta$ . $\sin \theta = \frac{\sqrt{x^2 - 9}}{x}$ SOOOOOO our answer in terms of $x$ is

$\frac{\sqrt{x^2 - 9}}{x} . \frac{1}{9} + C$ = $\frac{\sqrt{x^2 - 9}}{9x} + C$

now this is my answer and it seems correct. can anyone see anything wrong?

EDIT: found what was wrong and is now fixed

Hancock · « **Reply #48 on:** December 27, 2012, 09:50:54 pm »

Quote from: Nagisa on December 27, 2012, 09:41:54 pm

Evaluate the integral $\int \frac{1}{x^2\sqrt{x^2 - 9}} \ dx$

$let \ x = 3\sec \theta \ \ \ \ \ \ so \ dx = 3\sec \theta \tan \theta \ d\theta$

now apply back to the question.. specifically the $\sqrt{x^2 - 9}$

$\sqrt{x^2 - 9}$ = $\sqrt{9\sec^2\theta - 9}$ = $\sqrt{9(\sec^2\theta - 1)$ = $\sqrt{9\tan^2\theta}$ = $3 |\tan \theta| = 3\tan \theta$

$\therefore \int \frac{3\sec \theta \tan \theta}{9\sec^2\theta . 3\tan \theta} \ d \theta$ = $\frac{1}{3} \int \frac{1}{\sec \theta} \ d \theta$ = $\frac{1}{3} \sin \theta + C$

but this in terms of $\theta$ and the question is in terms of $x$ ... So remember when when we let $x = 3\sec \theta$ if we rearrange it we get $\sec \theta = \frac{x}{3}$ now this means we can make a magical triangle.. see below.

(Image removed from quote.)

so now we can apply our integral that was in terms of $\theta$ . $\sin \theta = \frac{\sqrt{x^2 - 9}}{x}$ SOOOOOO our answer in terms of $x$ is

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

now this is my answer and it seems correct. can anyone see anything wrong?

Yep, you made an error.

$\therefore \int \frac{3\sec \theta \tan \theta}{9\sec^2\theta . 3\tan \theta} \ d \theta$ = $\frac{1}{3} \int \frac{1}{\sec \theta} \ d \theta$ = $\frac{1}{3} \sin \theta + C$

$\frac{3}{9*3}$ should be equal to $\frac{1}{9}$ not $\frac{1}{3}$

http://www.wolframalpha.com/input/?i=integrate+1%2F(x%5E2(sqrt(x%5E2-9))

Nagisa · « **Reply #49 on:** December 27, 2012, 09:54:41 pm »

Quote from: Hancock on December 27, 2012, 09:50:54 pm

Yep, you made an error.

$\therefore \int \frac{3\sec \theta \tan \theta}{9\sec^2\theta . 3\tan \theta} \ d \theta$ = $\frac{1}{3} \int \frac{1}{\sec \theta} \ d \theta$ = $\frac{1}{3} \sin \theta + C$

$\frac{3}{9*3}$ should be equal to $\frac{1}{9}$ not $\frac{1}{3}$

http://www.wolframalpha.com/input/?i=integrate+1%2F(x%5E2(sqrt(x%5E2-9))

if u see clearly at the end i multiply it by 1/3 so i brought it back dw

Planck's constant · « **Reply #50 on:** December 27, 2012, 10:01:53 pm »

Quote from: Nagisa on December 27, 2012, 09:54:41 pm

if u see clearly at the end i multiply it by 1/3 so i brought it back dam

Any chance of a 'thank you Hancock for pointing out my mistake' ?

Nagisa · « **Reply #51 on:** December 27, 2012, 10:05:52 pm »

Quote from: argonaut on December 27, 2012, 10:01:53 pm

Any chance of a 'thank you Hancock for pointing out my mistake' ?

bro if you look closely u'll notice that hancock edited his message, his original message which i quoted was him saying that i didnt take notice of a 1/3

and can you just get out of my threads if ur just gonna say stuff that doesnt help anyone

Hancock · « **Reply #52 on:** December 27, 2012, 10:34:35 pm »

Yeah I noticed the wrong thing the first time, but I changed it before Nagisa posted his message (I didn't get a notification that someone had posted in the thread).

Quote

$\therefore \int \frac{3\sec \theta \tan \theta}{9\sec^2\theta . 3\tan \theta} \ d \theta$ = $\frac{1}{3} \int \frac{1}{\sec \theta} \ d \theta$ = $\frac{1}{3} \sin \theta + C$

should read:

$\therefore \int \frac{3\sec \theta \tan \theta}{9\sec^2\theta . 3\tan \theta} \ d \theta$ = $\frac{1}{9} \int \frac{1}{\sec \theta} \ d \theta$ = $\frac{1}{9} \sin \theta + C$

Argonaut: while you don't like Nagisa and you're sticking up for me (cheers btw), hostile attacks aren't really doing much to "quench the fire" (so to speak) that is brewing between you two.

Nagisa: Myself and others are helping you with issues you are having with maths on our own accord. While we are happy to do it, some gratitude does go a long way. You didn't see that $\frac{3}{3*9} = \frac{1}{9}$ which does cause you to arrive at the right answer.

Nagisa · « **Reply #53 on:** December 27, 2012, 10:43:48 pm »

Quote from: Hancock on December 27, 2012, 10:34:35 pm

Yeah I noticed the wrong thing the first time, but I changed it before Nagisa posted his message (I didn't get a notification that someone had posted in the thread).

Argonaut: while you don't like Nagisa and you're sticking up for me (cheers btw), hostile attacks aren't really doing much to "quench the fire" (so to speak) that is brewing between you two.

Nagisa: Myself and others are helping you with issues you are having with maths on our own accord. While we are happy to do it, some gratitude does go a long way. You didn't see that $\frac{3}{3*9} = \frac{1}{9}$ which does cause you to arrive at the right answer.

yeah this is true and i know i asked people to post their opinions. so im very happy that u help, i just dont like argonaut comin in and bein a smartass at me. there is no fire between us hes just being immature.

Nagisa · « **Reply #54 on:** December 27, 2012, 10:48:26 pm »

Need help with this one lol

evaluate $\int \frac{x^3}{\sqrt{x^2+9}} \ dx$ it's same method as last one just cant get it to fall out

Hancock · « **Reply #55 on:** December 27, 2012, 10:59:25 pm »

Quote from: Nagisa on December 27, 2012, 10:48:26 pm

Need help with this one lol

evaluate $\int \frac{x^3}{\sqrt{x^2+9}} \ dx$ it's same method as last one just cant get it to fall out

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

Let $u = x^2 + 9$

$\frac{du}{dx} = 2x$

$\int \frac{x^3}{\sqrt{x^2+9}} \ dx = \frac{1}{2}\int \frac{x^2}{\sqrt{u}} \ dx$

$x^2 = u - 9$

$\frac{1}{2}\int (u-9)u^{-0.5} dx$

$\frac{1}{2}\int u^{\frac{1}{2}}-9u^{\frac{-1}{2}} dx$

$\frac{1}{2}(\frac{2}{3}*u^{\frac{3}{2}} - 18u^{\frac{1}{2}})+C$

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

Hancock · « **Reply #56 on:** December 27, 2012, 11:05:06 pm »

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

Nagisa · « **Reply #57 on:** December 27, 2012, 11:13:16 pm »

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

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

yea ive tried $x = 3\tan \theta$

Hancock · « **Reply #58 on:** December 27, 2012, 11:20:49 pm »

I'll give it a shot with that one soon.

b^3 · « **Reply #59 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 $x\in [-\frac{\pi}{2},\frac{\pi}{2}]$

EDIT2: Made a small error, see following posts.

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