How would you solve something like this?

How would you solve something like this?

Other urls found in this thread:

m.wolframalpha.com/input/?i=integrate sin(x)/(1+cos^2(x))&x=0&y=0
twitter.com/NSFWRedditVideo

FUcking pethagoras, you pleb. for reals. that's all the hint youre getting. do your own fucking homework.

>pethagoras
how the fuck would that help anything

I think the denominator can be substituted with something like sin^2(u) or something. There's a formula.

Gimme a sec

Are you supposed to simply solve the integral and then subtracting both sides of the equal sign with pi/4 or am i missing something?

By posting it on Sup Forums

Substitute t=cos(u)

What do you do with the PI/4 - x part? I have never seen such an integral before

learn to use the internet, or get a fucking book with integration tables m.wolframalpha.com/input/?i=integrate sin(x)/(1+cos^2(x))&x=0&y=0

what about the left side of the equation???

pi/4 - x is equivalent to that definite integral OP, solve the integral, get the equivalence. Use the domain of x given to then narrow down answer.

Alright, thank you

what about the left part? thats like 5th grade maths just get x on one side and the numbers on the other.
If your question is why you don't have to integrate it pi/4 is a constant and the integral is du and not dx

this is wrong

see

answer is pi/4 i just did it

proof?

x=2460
thank me later fag

>x=2460
>x not within domain of [ 0, pi/3 ]
lol

yes, i was playing vidya so i couldnt write proof but this i s how i got it
pi/4 - x = S(0->pi) sin(u)du/(1+cos^2(u))

let V = 1 + cos^2(u)
dV = -2sin(u) du

(-1/2) S(0->pi)(1/V) dV
=
(-1/2)ln|1/(1+cos^2u)^1/2 (from 0 ->pi) = ln(1/2) - ln(1/2) = 0

so, 0 = pi/4 - x

x = pi/4
Sorry for the shitty notations lol

>(-1/2)ln|1/(1+cos^2u)^1/2 (from 0 ->pi) = ln(1/2) - ln(1/2) = 0


ignore the -1/2 at the beginning, I didn't mean to type it at the beginning since I transfered it over to the exponent anyway

THANK YOU user!!!!

No problem lol

dude you literally saved me from 6k in additional debt :DDD

OP that user fucked up.

if V = 1 + cos^2(u)
then dV = -2cos(u)sin(u)

I'll help with the problem give me a sec.

oh shit i did, felt i made a mistake somewhere'

do what will type

I hate math

hey i dont want to be mean or anything OP, but this is literally calc 1 tier stuff, and the fact that you asked about what to do with the left side makes me question whether or not you have the required skills to progress. This question is rather tame, and it will honestly get only more and more difficult.

Fuck man, I like it and I still hate it.

i love math but i'm shit at it

That's why I hate it. I like it because sometimes I'll do it right and feel like I've solved the Universe

x = pi/4
you substitute z = cos(u), which means dz = -sin(u)du, so the sin(u)du cancels out, leaving you with the integral from 1 to -1 of dz/(1+z^2), which you should be able to solve.

I'd pull out my calculator and ask Wong what the answer is.

back

Final answer is x=0. Haven't done calc in a while but I'm pretty confident of this solution. Learn your trig functions.

cos(0) = 1, and arctan(1)=pi/4, so it should be x = pi/4 right?

>solving math on lined paper

>not using the lines on lined paper
REEEEEEEEEEÈ

I was wrong, but you're wrong too. I got x= -pi/4, but that's out of scope. Can arctan() return values outside of [-pi/4, pi/4] ?

yeah I normally did math on blank paper, so fuck me :D

yeah, the limits to infinity and -infinity are pi/2 and -pi/2

Holy shit, I'm retarded. it's arccot() not arctan(). I'll fix it up.

okay yeah I think I got it. The integral solves to -pi/2 (since arctan(-1)-arctan(1) = -pi/4-pi/4 = -pi/2), so x becomes -pi/4

if you're not using Fourier series you done fucked up lad

Fucking forgot the oldest trick in the book

not OP but how did you rewrite the range?

My friend just asked: "Is there any intellectual content in Sup Forums?" I never thought I'd get to say yes.

Nevermind, I forgot you can't add constants on definite integrals. Fuck this question.