hi, when taking the integral of sec^3(theta)
how does the symbol turn from a positive to a negative
#exam tomorrow, integration by trig sub ( quick help)
exotic socket
echo loom
hi, when taking the integral of sec^3(theta)
how does the symbol turn from a po...
Probably for the same reason it integrates sec^3(theta) dx.
This doesn't look very carefully typed at all.
It's supposed to be the +, by the way.
Just worked that out.
exotic socket
hm, so it is supposed to be +
okay
good
i was going back and forth on my work
and checking mathway
i couldn't figure it out
echo loom
It's actually possible to evaluate int sec^3(x) dx yourself.
You start by taking it as sec^2(x) * sec(x) and using integration by parts.
Actually.
It might be the minus.
exotic socket
sec^2x= (1+tan^2x) * secx
echo loom
But it's the plus.
echo loom
sec^2x= (1+tan^2x) * secx
...no, you would integrate sec^2(x). Because the integral is just tan(x).
echo loom
Because what's d/dx tan(x)?
exotic socket
sec^2(x)
soo.. you would do tan(x) ln((sec(x)+tan(x))*
once you do integral sec^2(x)sec(x)
echo loom
...no?
Integration by parts. You integrate sec^2(x). You differentiate sec(x).
Remember? int u dv = uv - int v du?
exotic socket
yeah, im getting lost in the sauce
echo loom
Wait, how is this a trig sub though?
echo loom
Ah.
echo loom
That is, the integral of sec(x) dx?
exotic socket
what would the u- sub be for sec^2(x)sec(x)
exotic socket
well
YEAH
but you do
u=secx
du=secxtanx dx
dv=sec^2dx
v=tanx
ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
okay
needed to work it out
echo loom
Or you might use the DI method. In this case it's not wholly necessary because you're only doing one level of integration by parts.