MX2 Integration Marathon (1 Viewer)

idkkdi · Feb 1, 2021

Qeru said:
He assumed $sin^8$ is a constant so it's basically integrating $\sqrt{1-ax}$ where a is a constant

I know this bro/realised it

Qeru · Feb 1, 2021

Anyone wanna try my one lol

idkkdi · Feb 1, 2021

Qeru said:
Anyone wanna try my one lol

Not today chief

vernburn · Feb 1, 2021

idkkdi said:
Also just realised via checking calc that that expression doesn't have an actual integral.
Let me fix it
$\int \sqrt{1-\sin^4{x}} \text{dx}$

I may have stuffed up somewhere
$\int \sqrt{1-\sin^4{x}} dx$
$=\int \sqrt{(1-\sin^2x)(1+\sin^2x)} dx$
$=\int \sqrt{\cos ^2x(1+\sin^2x)} dx$
$=\int\cos x\sqrt{1+\sin^2x}dx$
let $\sin x = \tan\theta \implies \cos x dx = \sec^2\theta d\theta$ :
$= \int\sqrt{1+\tan^2\theta}\sec^2\theta d\theta$
$=\int\sec^3\theta d\theta$
$=\frac12\ln|\sec\theta+\tan\theta|+\frac12\sec\theta\tan\theta+C$
$=\frac12\ln|\sqrt{\sin^2x+1}+\sin x|+\frac{\sin x}{2}\sqrt{\sin^2x+1}+C$

vernburn · Feb 1, 2021

Qeru said:
Anyone wanna try my one lol

I'll have a go. Looks pretty tricky.

idkkdi · Feb 1, 2021

vernburn said:
I may have stuffed up somewhere
$\int \sqrt{1-\sin^4{x}} dx$
$=\int \sqrt{(1-\sin^2x)(1+\sin^2x)} dx$
$=\int \sqrt{\cos ^2x(1+\sin^2x)} dx$
$=\int\cos x\sqrt{1+\sin^2x}dx$
let $\sin x = \tan\theta \implies \cos x dx = \sec^2\theta d\theta$ :
$= \int\sqrt{1+\tan^2\theta}\sec^2\theta d\theta$
$=\int\sec^3\theta d\theta$
$=\frac12\ln|\sec\theta+\tan\theta|+\frac12\sec\theta\tan\theta+C$
$=\frac12\ln|\sqrt{\sin^2x+1}+\sin x|+\frac{\sin x}{2}\sqrt{\sin^2x+1}+C$

Alternative form or wrong desmos and symbolab time for u lol

idkkdi · Feb 1, 2021

vernburn said:
I'll have a go. Looks pretty tricky.

Med by day useless integrals by night. Why didn’t u take a maths degree or smth lol

vernburn · Feb 1, 2021

idkkdi said:
Med by day useless integrals by night. Why didn’t u take a maths degree or smth lol

Got nothing else to do rn

Qeru · Feb 1, 2021

vernburn said:
Got nothing else to do rn

Would you continue doing math after med?

vernburn · Feb 1, 2021

Qeru said:
Would you continue doing math after med?

prob won't do math again in two weeks time (apart from when helping my little sister lol)

notme123 · Feb 1, 2021

vernburn said:
$=\int\sec^3\theta d\theta$
$=\frac12\ln|\sec\theta+\tan\theta|+\frac12\sec\theta\tan\theta+C$
$=\frac12\ln|\sqrt{\sin^2x+1}+\sin x|+\frac{\sin x}{2}\sqrt{\sin^2x+1}+C$

how did you integrate sec cubed

idkkdi · Feb 1, 2021

notme123 said:
how did you integrate sec cubed

Split it

notme123 · Feb 1, 2021

idkkdi said:
Split it

yes ik the sec integrates to the first part but what about the second term
oh dont worry i found out how, by parts right

notme123 · Feb 1, 2021

this is probs a stupid q to ask and Ive asked this b4 but i dont think anyone answered but are these integrals just for fun, to deepen your understanding of the topic, or questions they could genuinely ask is a 4u hsc (without build up or previous questions).

Qeru · Feb 1, 2021

notme123 said:
this is probs a stupid q to ask and Ive asked this b4 but i dont think anyone answered but are these integrals just for fun, to deepen your understanding of the topic, or questions they could genuinely ask is a 4u hsc (without build up or previous questions).

They can ask everything here with build up only (or make it a last question like the 2014 hsc).

idekkedi · Feb 1, 2021

idkkdi said:
Split it

I guess you like being mentally challenged...

Qeru · Feb 1, 2021

idekkedi said:
I guess you like being mentally challenged...

How did you change the color lol

Directrix · Feb 1, 2021

Qeru said:
How did you change the color lol

Screenshot the pink I

then set that as ur pfp

Qeru · Feb 1, 2021

Directrix said:
Screenshot the pink I

then set that as ur pfp

Wrong account lol

Directrix · Feb 1, 2021

Qeru said:
Wrong account lol

U think it's me? Just telling you how to do that?

I can do urs if you like

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