MX2 Integration Marathon (1 Viewer)

notme123 · Feb 1, 2021

Qeru said:
Thats as simple as it gets I think (you can usually simplify if the inverse is in the brackets) also try using \cos^{-1}(x) to get $\cos^{-1}(x)$ imo looks a lot better

ok thanks

notme123 · Feb 1, 2021

Qeru said:
Really cool integral from stanford which a lot of people have probably seen before:

$\int_{0}^{\pi} \frac{2x \sin{x}}{3+\cos{2x}} dx$

lmao is the answer 0? If not, I screwed up

idkkdi · Feb 1, 2021

notme123 said:
lmao is the answer 0

yep

notme123 · Feb 1, 2021

idkkdi said:
yep

its not i think

Qeru · Feb 1, 2021

The answer is: $\frac{{\pi}^2}{4}$

notme123 · Feb 1, 2021

Qeru said:
The answer is: $\frac{\pi^2}{4}$

wait let me see my mistale

idkkdi · Feb 1, 2021

Qeru said:
The answer is: $\frac{{\pi}^2}{4}$

The graph on desmos looks like 0?

notme123 · Feb 1, 2021

Qeru said:
The answer is: $\frac{{\pi}^2}{4}$

OH IM SUCH A DUMB POO I WAS LITERALLY AT $\int \frac{\pi}{2}-\sin^{-1}(\tan(u))du$ pretend from -pi\4 to pi/4
the sin turns to 0 because its odd
and when i did the upper and lower bounds on (pi/2)u i plused not minused pls guys i swear im not crap

idkkdi · Feb 1, 2021

Qeru said:
The answer is: $\frac{{\pi}^2}{4}$

Wasn't that in MIt integration bee as well?

idkkdi · Feb 1, 2021

notme123 said:
OH IM SUCH A DUMB POO I WAS LITERALLY AT $\int \frac{\pi}{2}+\sin^{-1}(\tan(u)}du$

Suffering latex pain I c

idekkedi · Feb 1, 2021

idkkdi said:
mmm yes sexy latex completed

Sexier than you will ever be.

idkkdi · Feb 1, 2021

idekkedi said:
Sexier than you will ever be.

This is clearly @Velocifire @Directrix ‘s new alt slandering me wtf.

YonOra · Feb 1, 2021

Who needs the hub when you got maths?

Directrix · Feb 1, 2021

@Velocifire @Directrix ‘s new alt slandering me wtf.
[/QUOTE]
I would do

KidKid instead

notme123 · Feb 1, 2021

Qeru said:
Really cool integral from stanford which a lot of people have probably seen before:

$\int_{0}^{\pi} \frac{2x \sin{x}}{3+\cos{2x}} dx$

$=\int_{0}^{\pi} \frac{2x \sin{x}}{3+(2\cos^2{x}-1)} dx$
$=\int_{0}^{\pi} \frac{x \sin{x}}{1+\cos^2{x}} dx$
$\tan(u)=\cos(x)$
$dx=-\frac{\sec^2(u)du}{\sin(x)}$
$=-\int_{\frac{\pi}{4}}^{-\frac{\pi}{4}} \frac{x\sec^2(u)}{1+\tan^2(u)}du$
$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} xdu$
$=\int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^{-1}(\tan(u))du$
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi}{2}-\sin^{-1}(u)du$ yes I know I need to explain myself when using this identity and also the sin^-1 u is 0 because odd with opposite bounds
$=\frac{\pi}{4}*\frac{\pi}{2}-(-\frac{\pi}{2}*\frac{\pi}{4})$
$=\frac{\pi^2}{4}$

notme123 · Feb 1, 2021

idkkdi said:
more integrals

$\int \frac{\sin{\sqrt{x}+a}(e^{\sqrt{x}})}{\sqrt{x}}$

really?
$-2\cos(\sqrt{x})+2ae^{\sqrt{x}}+c$

Qeru · Feb 1, 2021

Qeru said:
Really cool integral from stanford which a lot of people have probably seen before:

$\int_{0}^{\pi} \frac{2x \sin{x}}{3+\cos{2x}} dx$

Or we can use the king Rule:

$I=\int_{0}^{\pi} \frac{2x \sin{x}}{3+\cos{2x}} dx$ (1)

$I=-\int_{\pi}^{0} \frac{2 (\pi-x) \sin{x}}{3+\cos{2x}} dx \quad \text{by letting} \quad u=\pi-x$

$I=\int_{0}^{\pi} \frac{2 (\pi-x) \sin{x}}{3+\cos{2x}} dx$ (2)

Add 1 and 2 to get:

$I=\int_{0}^{\pi} \frac{2\pi\sin{x}}{3+\cos{2x}} dx$

$I=2\pi\int_{0}^{\pi} \frac{\sin{x}}{2\cos^2{x}+2} dx$

$I=\pi\int_{0}^{\pi} \frac{\sin{x}}{\cos^2{x}+1} dx$

$I=-\pi\left[\tan^{-1}(\cos{x})\right]_{0}^{\pi}$

$I=\frac{{\pi}^2}{4}$

vernburn · Feb 1, 2021

Qeru said:
Or we can use the king Rule:

$I=\int_{0}^{\pi} \frac{2x \sin{x}}{3+\cos{2x}} dx$ (1)

$I=-\int_{\pi}^{0} \frac{2 (\pi-x) \sin{x}}{3+\cos{2x}} dx \quad \text{by letting} \quad u=\pi-x$

$I=\int_{0}^{\pi} \frac{2 (\pi-x) \sin{x}}{3+\cos{2x}} dx$ (2)

Add 1 and 2 to get:

$I=\int_{0}^{\pi} \frac{2\pi\sin{x}}{3+\cos{2x}} dx$

$I=2\pi\int_{0}^{\pi} \frac{\sin{x}}{2\cos^2{x}+2} dx$

$I=\pi\int_{0}^{\pi} \frac{\sin{x}}{\cos^2{x}+1} dx$

$I=-\pi\left[\tan^{-1}(\cos{x})\right]_{0}^{\pi}$

$I=\frac{{\pi}^2}{4}$

Exactly how I would have done it. Great minds think alike!

notme123 · Feb 1, 2021

Qeru said:
Or we can use the king Rule:

$I=\int_{0}^{\pi} \frac{2x \sin{x}}{3+\cos{2x}} dx$ (1)

$I=-\int_{\pi}^{0} \frac{2 (\pi-x) \sin{x}}{3+\cos{2x}} dx \quad \text{by letting} \quad u=\pi-x$

$I=\int_{0}^{\pi} \frac{2 (\pi-x) \sin{x}}{3+\cos{2x}} dx$ (2)

Add 1 and 2 to get:

$I=\int_{0}^{\pi} \frac{2\pi\sin{x}}{3+\cos{2x}} dx$

$I=2\pi\int_{0}^{\pi} \frac{\sin{x}}{2\cos^2{x}+2} dx$

$I=\pi\int_{0}^{\pi} \frac{\sin{x}}{\cos^2{x}+1} dx$

$I=-\pi\left[\tan^{-1}(\cos{x})\right]_{0}^{\pi}$

$I=\frac{{\pi}^2}{4}$

ik this trick is too advanced for me but pls explain the last step to inverse tan cos
OH WAIT IM SPECIAL DONT WORRY LOL

Qeru · Feb 1, 2021

notme123 said:
ik this trick is too advanced for me but pls explain the last step to inverse tan cos

So it's jsut using the result $\int \frac{f'(x)}{1+(f(x))^2} dx =\tan^{-1}(f(x))+C$ If you want you could do $u=\cos(x)$ and proceed from there.

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