Question 1e... What did you do? (1 Viewer)

[biopia]

Hey everyone.
I'd be interested to know what people did for question 1e...
If I remember correctly, it was the integral of:
1/[x²√(1+x²)]

I am hoping I did it right :S
I made the substitution of x = tanθ...
If I did it correctly, it worked out to be the integral of cos³x/sin²x
Don't shoot me down lol! I am pretty fragile after that exam hahaha

I forget the limits lol.

 

[Dumbledore]

factorise x^2 out of the square root to get:

1/(x^3 * sqrt(1/x^2 + 1))

then let u = 1/x^2 so the 1/x^3 will cancel

OR

x = tan@ and u should end up with cos@/sin^2@ i think

 

[tommykins]

The answer is -sqrt(1+x^2)/x if you're wondering.

 

[khorne]

Not a hard integral:
sub: x = tanu

dx=du sec^2u, so rt(x^2 +1) = rt(sec^2(u)) = sec(u), and u = tan^-1(x), which then ends up as:

integral of cot(u)cosec(u) = -cosec(u), and sub back, you get - sqrt(x^2 +1)/x

 

[untouchablecuz]

x=tan(u)

end up with

∫(sinu)^-2cosudu

 

[Timothy.Siu]

yeah subbed in tan

but i got cosx/sin^2 x not what u got lol

 

[les coccinelles]

i got the same as untouchablecuz, but i integrated wrong lol.

 

[biopia]

Lol, seems as if I fucked up the question lol. Not too much though, so hopefully, I'l at least get 2/4 for that answer.

The only thing I forgot to do was replace the dx... Oh well haha.

 

[andrew234]

uh i did it differently

i let u = (1+x^2)^1/2
and then after a few linese of working out , i got

2/(u-1)(u+1)
and had to use partial fractions

can anyone confirm if this way is correct as well

in the end i got something like
-ln3 - ln(2^1/2-1) + ln(2^1/2 +1)

 

[00iCon]

ARGH! I did it by partial fractions and couldn't finish... then i tried continuing by subbing x=sin@ but that failed...
IMO, the hardest question in the whole exam was Q1a...

 

[biopia]

It's funny how sometimes the easieest questions can be so hard :S
One of my mates in my class couldn't get 1a either.
Pretty sure it was just a substitution of u=lnx so du=dx.x which just made it in the integral of u.
Exam conditions make everything come undone. Like seriously, who forgets to sub back in for dx!!!!! Argh...
I was used to integration being a little easier than that lol.

 

[tommykins]

sub in limits with the answer we gave you and see if it equates to urs

 

[Puttah]

I did that substitution at first too, but you end up with (u^2-1)^(-3/2) which seems pretty damn hard to integrate on its own...
The proper sub is x=tanu

 

[Monsterman]

uh i did it differently

i let u = (1+x^2)^1/2
and then after a few linese of working out , i got

2/(u-1)(u+1)
and had to use partial fractions

can anyone confirm if this way is correct as well

in the end i got something like
-ln3 - ln(2^1/2-1) + ln(2^1/2 +1)

are you sure its.. 2/(u-1)(u+1)...

 

[gurmies]

I think it was sqrt[2]-2/sqrt[3]

 

[chantszhin]

Well there's always multiples solutions to integration questions!!
I did x=tan@

 

[Aerath]

Well there's always multiples solutions to integration questions!!
I did x=tan@

Not really if there's borders.

 

[lychnobity]

That's what I did (with limits in the exam though)

 

[Blaz-357]

that's what i did (with limits in the exam though)

+1