#Trig Sub
86 messages · Page 1 of 1 (latest)
You want the denominator to look like 1 - (sin x)^2 inside the square root
Then you can replace that with (cos x)^2
You have a 16 instead of a 1, so your target is going to be 16 - 16(sin x)^2, and then you'll be able to common factor the 16.
?
Yeah, either works, but you get to choose, so you might as well choose the positive one
I get to choose the sign for the substitution?
you get to choose the entire substitution. You could choose x = 4 cos(theta) too.
don't interrupt. open your own thread
when you did the substitution, you didn't replace the bounds
Nice. Looks good.
(It's long and a bit intricate, so I don't guarantee that it's all correct, but the ideas and process are definitely right, and I think the answer is correct.)
I tried being as precise as possible
Hopefully it's right
I'm doing another one right now
But I'm stuck
convert to sines and cosine
?
go to sines and cosines
?
yeah
take a moment and think about this one
it's not too hard, but it's also not automatic
What would've been an alternative to this?
you can sub u = sin(theta), and the cos(theta)d(theta) becomes du
so it's just integral of 1/u^2 du
yes
I get
seems good
Isn’t there a way I can rewrite the arctan
So that I don’t have any trig functions in my final answer
This is weird though
a little bit, yeah
,w $\int \frac{dx}{e^x\sqrt{4+e^{2x}}}$
I don't know how to use this bot 😭
yeah, draw a triangle
Yeah
sorry, brain fart
Wait, so how should I proceed?
suppose you have cos(arcsin(x))
you draw a right triangle with opposite x, hypotenuse 1
because that's what arcsin(x) means
then you pythagorean theorem and find cos
This correct?
Different answer here…
Do you know where I made my mistake?
Oh, never mind I got it
Thanks for your help