/sci/ - Science & Math

[Anonymous No.1953976]

Hey /sci/ I have a problem with integration by parts.
<span class="math">\int\frac{\cos x}{\sin x}\,dx = |u = \frac{1}{\sin x},\;dv = \cos x\,dx | = \frac{\sin x}{\sin x} - \int\frac{(-\cos x)\sin x}{\sin^2 x}\,dx = 1 + \int\frac{\cos x}{\sin x}\,dx = 2 + \int\frac{\cos x}{\sin x}\,dx = \dots = n + \int\frac{\cos x}{\sin x}\,dx \iff 0=1=2=\dots=n[/spoiler]

[Anonymous No.1953980]

2 1/2 sage

[Anonymous No.1953990]

Troll Math here? Oh you.

[Anonymous No.1954022]

bump fags

[Anonymous No.1954136]

bump

[Anonymous No.1954142]

even for troll math, this is really bad. most people are familiar with constants of integration

[Anonymous No.1954145]

<div class="math">\int \frac{cos(x)}{sin(x)}dx=ln(sin(x))+C</div>

oh god what is this I am not good with immediate integrals

[Anonymous No.1954209]

>>1954145
You've just spoiled my stupid problem. And what about 0=1=2=...=n?

[Anonymous No.1954226]

>>1954145
>mfw you are wrong
<div class="math">\int \frac{\cos(x)}{\sin(x)}dx=\ln | \sin(x) |+C</div>

[Anonymous No.1954244]

>>1954209
<span class="math">\int\frac{\cos x}{\sin x}\,dx=\dots=n+\int\frac{\cos x}{\sin x}\,dx = \ln(\sin x) + c \ffi n + \ln(\sin x) + c = \ln(\sin x) + c \implies \forall n \in \Nu n = 0[/spoiler]

[Anonymous No.1954248]

>implying sin(x)/cos(x) isnt 1/tan(x)

[Anonymous No.1954251]

>>1954244
kewl trawlin broh

[Anonymous No.1954261]

<span class="math">\int\frac{\cos x}{\sin x}\,dx=\dots=n+\int\frac{\cos x}{\sin x}\,dx,\;\int\frac{\cos x}{\sin x}\,dx=\ln(|\sin x|)+c\implies n+\ln(|\sin x|)+c=\ln(|\sin x|) + c\implies \forall n \in \Nu n = 0[/spoiler]

[Anonymous No.1954265]

>>1954261
>implying

[Anonymous No.1954269]

Unknown control sequence '\implies'
>mfw you can't imply

[Anonymous No.1954274]

>uses integration by parts when he could just substitute cos(x)
>laughing_girls.jpg