MX2 Integration Marathon (4 Viewers)

idkkdi · Jan 30, 2021

vernburn said:
Okay here’s my answer but it looks very VERY wrong. Don’t mock me!

Definitely could simplify inside brackets but it looks decently legit

idkkdi · Jan 30, 2021

@Qeru latex a solution lol

vernburn · Jan 30, 2021

vernburn said:
Okay here’s my answer but it looks very VERY wrong. Don’t mock me!

It’s wrong. Just realised I stuffed up in my solution

.

Qeru · Jan 30, 2021

idkkdi said:
@Qeru latex a solution lol

Be back in 30mins once I type this up lol

vernburn · Jan 30, 2021

Attempt #2 looks even worse lol.

notme123 · Jan 30, 2021

Something's very wrong, I can feel it.
$\int{ \sqrt{\csc^2{x}+\cot^2{x}} dx =\int{ \frac{\sqrt{1+\cos^2(x)}}{\sin(x)}dx$
$u^2=1+\cos^2x$
$2udu=-2\cos(x)*\sin(x)dx$
$dx=-\frac{u}{\sin(x)*\cos(x)}du$
$-\int{\frac{u^2}{\sin^2(x)*\cos(x)}du$
$=-\int{\frac{u^2}{\sqrt{u^2-1}*(2-u^2)}du$
$w^2=u^2-1$
$du=\frac{w}{u}dw$
$-\int{\frac{u^2}{w*(1-w^2)}*\frac{w}{u}dw$
$=\int{\frac{\sqrt{w^2+1}}{{(w^2-1)}}dw$
$w=\tan(v)$
$dw=\sec^2(v)dv$
$\int{\frac{\sec^3(v)}{{(\tan^2v-1)}}dv$
$= \int \frac{1}{\cos{v}(\sin^2{v}-\cos^2{v})} dv$
...

Qeru · Jan 30, 2021

Qeru said:
Yup nice now for a nasty one:

$\int{ \sqrt{\csc^2{x}+\cot^2{x}} dx$

Let $\cos{x}=\tan{u} \implies-\sin{x} dx=\sec^2{u}du$

$I=-\int \sqrt{\frac{1+\tan^2{u}}{(1-\tan^2{u})^2}} \times \sec^2{u}du \quad \text{since} \quad \sin{x}=\sqrt{1-\tan^2{u}}$

$I=-\int \frac{\sec^3{u}}{1-\tan^2{u}} du$

$I=-\int \frac{\frac{1}{\cos^3{u}}}{1-\frac{\sin^2u}{\cos^2{u}} du$

$I=- \int \frac{1}{\cos{u}(\cos^2{u}-\sin^2{u})} du$

Let $s=\sin{u} \implies ds=\cos{u} du \implies \frac{1}{\cos{u}}ds=du$

$I=-\int \frac{1}{(1-s^2)(1-2s^2)} du$

$I=\int \frac{1}{(s-1)(s+1)(1-\sqrt{2}s)(1+\sqrt{2}s)} ds$

$I=\int \frac{1}{(s-1)(s+1)(1-\sqrt{2}s)(1+\sqrt{2}s)} ds$

$I=\int \frac{1}{2}\left(\frac{1}{s+1}-\frac{1}{s-1}\right)+\frac{2}{(1-\sqrt{2}s)(1+\sqrt{2}s)}ds$ Ill save the partial fractions working.

$I=\int \frac{1}{2}\left(\frac{1}{s+1}-\frac{1}{s-1}\right)+\left(\frac{1}{1-\sqrt{2}s}+\frac{1}{1+\sqrt{2}s}\right)ds$

$I=\frac{1}{2} \ln\left(\frac{s+1}{s-1}\right)+\frac{1}{\sqrt{2}} \ln {\left(\frac{1+\sqrt{2}s}{1-\sqrt{2}s}\right)} +C$

Drawing up a triangle we find: $s=\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}$

So:

$I=\frac{1}{2} \ln\left(\frac{\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}+1}{\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}-1}\right)+\frac{1}{\sqrt{2}} \ln {\left(\frac{1+\sqrt{2}\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}}{1-\sqrt{2}\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}}\right)} +C$

I could have very easily made a mistake whether in latex or just trying to evaluate to integral so let me know if you find any.

Edit, fixed sign mistakes.

Qeru · Jan 30, 2021

If anyone is bothered see if my result matches with vernburn's lol

notme123 · Jan 30, 2021

are these questions they could straight up ask in an HSC exam? Or is this just for the lols?

vernburn · Jan 30, 2021

Qeru said:
Let $\cos{x}=\tan{u} \implies-\sin{x} dx=-\sec^2{u}du$

$I=\int \sqrt{\frac{1+\tan^2{u}}{(1-\tan^2{u})^2}} \times \sec^2{u}du \quad \text{since} \quad \sin{x}=\sqrt{1-\tan^2{u}}$

$I=\int \frac{\sec^3{u}}{1-\tan^2{u}} du$

$I=\int \frac{\frac{1}{\cos^3{u}}}{1-\frac{\sin^2u}{\cos^2{u}} du$

$I=\int \frac{1}{\cos{u}(\cos^2{u}-\sin^2{u})} du$

Let $s=\sin{u} \implies ds=-\cos{u} du \implies -\frac{1}{\cos{u}}ds=du$

$I=-\int \frac{1}{(1-s^2)(1-2s^2)} du$

$I=\int \frac{1}{(s-1)(s+1)(1-\sqrt{2}s)(1+\sqrt{2}s)} ds$

$I=\int \frac{1}{(s-1)(s+1)(1-\sqrt{2}s)(1+\sqrt{2}s)} ds$

$I=\int \frac{1}{2}\left(\frac{1}{s+1}-\frac{1}{s-1}\right)+\frac{2}{(1-\sqrt{2}s)(1+\sqrt{2}s)}ds$ Ill save the partial fractions working.

$I=\int \frac{1}{2}\left(\frac{1}{s+1}-\frac{1}{s-1}\right)+\left(\frac{1}{1-\sqrt{2}s}+\frac{1}{1+\sqrt{2}s}\right)ds$

$I=\frac{1}{2} \ln\left(\frac{s+1}{s-1}\right)+\frac{1}{\sqrt{2}} \ln {\left(\frac{1+\sqrt{2}s}{1-\sqrt{2}s}\right)} +C$

Drawing up a triangle we find: $s=\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}$

So:

$I=\frac{1}{2} \ln\left(\frac{\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}+1}{\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}-1}\right)+\frac{1}{\sqrt{2}} \ln {\left(\frac{1+\sqrt{2}\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}}{1-\sqrt{2}\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}}\right)} +C$

I could have very easily made a mistake whether in latex or just trying to evaluate to integral so let me know if you find any

Nice. I did the same initial sub then got to $I=\int \frac{1}{\cos{u}(\cos^2{u}-\sin^2{u})} du$ and used a t-sub. Your method is much better and I suspect I have made an error somewhere.

Directrix · Jan 30, 2021

I'm afraid I have to step in here.

What is 1+1

notme123 · Jan 30, 2021

Directrix said:
I'm afraid I have to step in here.

What is 1+1

I think you need uni level maths to do that.

notme123 · Jan 30, 2021

Qeru said:
Let $\cos{x}=\tan{u} \implies-\sin{x} dx=-\sec^2{u}du$

$I=\int \sqrt{\frac{1+\tan^2{u}}{(1-\tan^2{u})^2}} \times \sec^2{u}du \quad \text{since} \quad \sin{x}=\sqrt{1-\tan^2{u}}$

$I=\int \frac{\sec^3{u}}{1-\tan^2{u}} du$

$I=\int \frac{\frac{1}{\cos^3{u}}}{1-\frac{\sin^2u}{\cos^2{u}} du$

$I=\int \frac{1}{\cos{u}(\cos^2{u}-\sin^2{u})} du$

Let $s=\sin{u} \implies ds=-\cos{u} du \implies -\frac{1}{\cos{u}}ds=du$

$I=-\int \frac{1}{(1-s^2)(1-2s^2)} du$

$I=\int \frac{1}{(s-1)(s+1)(1-\sqrt{2}s)(1+\sqrt{2}s)} ds$

$I=\int \frac{1}{(s-1)(s+1)(1-\sqrt{2}s)(1+\sqrt{2}s)} ds$

$I=\int \frac{1}{2}\left(\frac{1}{s+1}-\frac{1}{s-1}\right)+\frac{2}{(1-\sqrt{2}s)(1+\sqrt{2}s)}ds$ Ill save the partial fractions working.

$I=\int \frac{1}{2}\left(\frac{1}{s+1}-\frac{1}{s-1}\right)+\left(\frac{1}{1-\sqrt{2}s}+\frac{1}{1+\sqrt{2}s}\right)ds$

$I=\frac{1}{2} \ln\left(\frac{s+1}{s-1}\right)+\frac{1}{\sqrt{2}} \ln {\left(\frac{1+\sqrt{2}s}{1-\sqrt{2}s}\right)} +C$

Drawing up a triangle we find: $s=\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}$

So:

$I=\frac{1}{2} \ln\left(\frac{\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}+1}{\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}-1}\right)+\frac{1}{\sqrt{2}} \ln {\left(\frac{1+\sqrt{2}\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}}{1-\sqrt{2}\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}}\right)} +C$

I could have very easily made a mistake whether in latex or just trying to evaluate to integral so let me know if you find any

with the s sub why when differentiated sin is negative cos

Qeru · Jan 30, 2021

notme123 said:
with the s sub why when differentiated sin is negative cos

Pretend you didnt see that

. Also yes rookie mistake and fixed.

notme123 · Jan 30, 2021

Qeru said:
Pretend you didnt see that . Also yes rookie mistake and fixed.

It's ok it's amazing you even were able to get to that point. I saw another one but I think I may be wrong because you and vern got to the same point but also the first sub why is the differential of tan -sec squared
Dw you fixed it

Qeru · Jan 30, 2021

notme123 said:
It's ok it's amazing you even were able to get to that point. I saw another one but I think I may be wrong because you and vern got to the same point but also the first sub why is the differential of tan -sec squared
Dw you fixed it

Yeah honestly don't know what I was thinking, ironically the final answer is correct since the two negatives (which were both mistakes) cancel out.

Qeru · Jan 30, 2021

notme123 said:
Something's very wrong, I can feel it.
$\int{ \sqrt{\csc^2{x}+\cot^2{x}} dx =\int{ \frac{\sqrt{1+\cos^2(x)}}{\sin(x)}dx$
$u^2=1+\cos^2x$
$2udu=-2\cos(x)*\sin(x)dx$
$dx=-\frac{u}{\sin(x)*\cos(x)}du$
$-\int{\frac{u^2}{\sin^2(x)*\cos(x)}du$
$=-\int{\frac{u^2}{\sqrt{u^2-1}*(2-u^2)}du$
$w^2=u^2-1$
$du=\frac{w}{u}dw$
$-\int{\frac{u^2}{w*(1-w^2)}*\frac{w}{u}dw$
$=\int{\frac{u}{{(w^2-1)}}dw$

Yeah everything is good till the last step when you have to convert the u to a w so you will get a square root again.

notme123 · Jan 30, 2021

Qeru said:
Yeah honestly don't know what I was thinking, ironically the final answer is correct since the two negatives (which were both mistakes) cancel out.

You're invincible at maths. Even when you make mistakes it's not really a mistake and works out in your favor.

notme123 · Jan 30, 2021

Qeru said:
Yeah everything is good till the last step when you have to convert the u to a w so you will get a square root again.

i just gave up at the end cuz i knew it was gonna look ugly
actually i couldve subbed in w = tanv or something

Qeru · Jan 30, 2021

notme123 said:
i just gave up at the end cuz i knew it was gonna look ugly
actually i couldve subbed in w = tanv or something

Yep and you would have got what Vern and I got. The sub $\cos{x}=\tan{u}$ does your two subs in one.

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