MX2 Integration Marathon (1 Viewer)

idkkdi · Feb 1, 2021

idkkdi said:
Real integral
$\int \sqrt {1-sin^8{x}}dx$

Qeru said:
Ok anyone want to try this (stumped me lol):

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

In the meantime try my one

vernburn · Feb 1, 2021

idkkdi said:
Real integral
$\int \sqrt {1-sin^8{x}}dx$

Easy!
$-\frac{2(1-sin^8x)\sqrt{1-sin^8x}}{3sin^8}+C$

idkkdi · Feb 1, 2021

vernburn said:
Easy!
$-\frac{2(1-sin^8x)\sqrt{1-sin^8x}}{3sin^8}+C$

Where's the x

idekkedi · Feb 1, 2021

idkkdi said:
Where's the x

Your x is gone and don't ask me y!

Qeru · Feb 1, 2021

vernburn said:
Easy!
$-\frac{2(1-sin^8x)\sqrt{1-sin^8x}}{3sin^8}+C$

loooool, thats big brain

idkkdi · Feb 1, 2021

vernburn said:
Easy!
$-\frac{2(1-sin^8x)\sqrt{1-sin^8x}}{3sin^8}+C$

Also, pretty sure that ain't it

idkkdi · Feb 1, 2021

Qeru said:
loooool, thats big brain

U guys actually trolling me what's going on wtf

vernburn · Feb 1, 2021

idkkdi said:
U guys actually trolling me what's going on wtf

You wrote $sin^8$ not $\sin^8$

Qeru · Feb 1, 2021

idkkdi said:
U guys actually trolling me what's going on wtf

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

idkkdi · Feb 1, 2021

idkkdi said:
U guys actually trolling me what's going on wtf

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

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)

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