#help
77 messages · Page 1 of 1 (latest)
First part of second page is wrong I think
how @compact cedar
U cant square the bottom and top
It's not the same
Atleast that's what I'm assuming you did
ðŸ˜ðŸ˜
Brother U can't square the top and bottom of a fraction and it's the same
Like 3/4 x 3/4 is not the same as 3/4
^^^
Trig sub
I was thinking of using sum else tho icl
4sinhu
Or would that be wrong
wrong
i havent learned hyperbolic
Wait would it be 1/4 sinhu
this is how R2drew2 did it
Silly me
yeh it's a standard integral
just parts + arctan formula
bro icl this question is making me pregnant imma try this tmr
By parts?
Ahh What is a standard integral
ones you should know
What’s the answer
Oh shet I js make the sub up as i go
Oh I answered this one
Oh so since it wants the answer to have arctan u use tan sub?
Since icl I would've done it a bit differently but I'm not sure it'd work
you rarely actually do the sub
I'm trying it with 1/4 sinhu is there a reason why it wouldn't work
yeh cus when you differentiate arsinh,arcosh and arcsin and arccos you get a square root on the denominator
Ohh so since it looks like an arctan u use tan sub?
but when you differentiate artanh and arctan you don't get a square root on the denominator
yeh
Ah I was js using anything good to knoa
Thanks bro
Also say if the trig integration looks like u can write it down from the equation sheet am I fine to do thst or should I show working for marks?
Guys
Question:
My working:
I made a division mistake here: underlined
@hollow arch
you're still tryna integrate this right?
Yh
So at the end of this you should get … - integral of 2x^2/(1+16x^2) dx
you can kind of see the top part is almost a multiple of the bottom part so it can be written as
integral of (2x^2 + 1/8)/(1+16x^2) - (1/8)/(1+16x^2) dx
And now the numerator of the first fraction is a multiple of the denominator so
It’s now the integral of 1/8 - (1/8)/1+16x^2 dx
Factor out the 1/8 and use standard result for arctan
I know it’s different to how you’ve done it but you HAVE to get used to using these standard results and applying them
@west hinge @fast hornet
Yeh that’s what I got
Noice
@west hinge I would use partial fraction for this