#now why

[tonycries]

∏_{j=1}^∞ cos(x/2j) = ∏_{j=1}^∞ [(1/2) * (cos(x/(2j-1)) + cos(x/(2j+1)))]

= ∏_{j=1}^∞ [1/2 * cos(x/(2j-1))] * ∏_{j=1}^∞ [1/2 * cos(x/(2j+1))]

= (1/2)^∞ * cos(x/1) * cos(x/3) * cos(x/5) *

∫_{0}^{π/2} x * ∏_{j=1}^∞ cos(x/2j) dx

= ∫_{0}^{π/2} x * (1/2)^∞ * cos(x/1) * cos(x/3) * cos(x/5) * ... dx

= (1/2)^∞ * ∫_{0}^{π/2} x * cos(x/1) * cos(x/3) * cos(x/5) * ... dx

cos(x/1) * cos(x/3) * cos(x/5) * ... = (1/2) * (sin(x) + sin(3x)/3 + sin(5x)/5 + ...)

∫_{0}^{π/2} x * cos(x/1) * cos(x/3) * cos(x/5) * ... dx

= (1/2)^∞ * ∫_{0}^{π/2} x * (sin(x) + sin(3x)/3 + sin(5x)/5 + ...) dx

= (1/2)^∞ * [(-x*cos(x))/1 + (1/3)*x*cos(3x) - (1/5)*x*cos(5x) + ...]_{0}^{π/2}

= (1/2)^∞ * [(-π/2)*cos(π/2)/1 + (1/3)*(π/2)*cos(3*(π/2)) - (1/5)*(π/2)*cos(5*(π/2)) + ...]

= (1/2)^∞ * [(π/2)/3 - (π/2)/5 + (π/2)/7 - ...]

= (1/2)^∞ * π/4 * [1/1 - 1/3 + 1/5 - ...]

1 - 1/3 + 1/5 - 1/7 + ... = π/4

Therefore,

∫_{0}^{π/2} x * ∏_{j=1}^∞ cos(x/2j) dx

= (1/2)^∞ * π/4

= 0

[06strawberrylemonade28]

so my brain has decided to be obsessed with beast boy and raven now... what the heck

if anyone has any bbrae or teen titans found family fic recs pLEASEEE send them my way

[meadowfics]

oh that’s not—

??

[foggynelsonarchive]

At least Foggy doesn't have student loans anymore.

[falsenote]

[sluttywonwoo]

Every time I find myself a little too delulu about mingyu and start planning out my bedtime scenarios i remind myself that he is a fully grown man with cock and balls and has had to clech with his core with all of his strength while on the toilet at least once in his life. Brings me back to reality asap

[kaz-lovebot]