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.