Veeky Forums can't even solve this Calculus 1 problem
Veeky Forums can't even solve this Calculus 1 problem
>homework
>he doesn't know pythagoras theorem
not a solution
Isn't it Calculus 2? I had the something like this on a recent exam only it was from 0 to 2pi and that equals 0, I think this one is pi^2/4?
anyway the principle is the same, you write the antiderivative of sin(x)/(1+(cos(x))^2) and integrate by parts
nope
Is it 0?
>calculus 1
top kek
i found 4 poles, with the residue theorem i get pi2/4 for your integral.
It's late so i'm not sure about my shit.
I have a feeling the answer will include a fuckton of terms.
Without evaluating the integral I mean.
Its 0
It isn't
singularity at z=pi tho
Take your pedophile cartoons back to and your homework back to .
yea, sorry, it's 2 AM here, it's pi^2/4?
The solution was just posted in another thread exactly like this one. It is pi^2/4. You make the substitution u=pi-x. I need to remember to use similar substitutions like this when dealing with trigonometric or other periodic integrals.
yep
I found this one much harder
sorry, I misstyped, It's supposed to be (x+1)^3 down
It doesn't claim to be a solution either you autistic fuck. do your own homework.
Calculate [math]\int \frac{\sin\,x}{1 \,+\, \left(\cos\, x\right)^2} \,\mathrm dx[/math] separately and use the result to integrate by parts.
Now back to the homework board.