Can someone spot where I went wrong?

>>8628558
derivative of sin is cos not (-cos).
didnt read the rest cause thats basic fuckup.

>>8628563
Thanks

>>8628568
np my friend.

>>8628568
wait im dumb lol
nevermoind me forget what i said.

>>8628568
you got that right the derivative of (-cos) is sin.

>>8628558
[math]\int \sin{2x} dx=-\frac{1}{2} \cos{2x}[/math]

>>8628585
but my dude where does he integrate sin(2x) ?

I say that its just -3xcos(2x)

because you dont write the limits in this function, it already is the integral. then integral of cos(2x) is 1/2sin(2x) and this with regards to 4 pi and 0 is still zero
so the answer should be -3xcos2x
amirite

>>8628585
Why is this?

>>8628558
but why do you have cos of pi in your result ? thats just minus 1 !

>>8628602
because of chain rule i guess

>>8628598
>but my dude where does he integrate sin(2x) ?

When he does integration by parts.

>>8628602
Come on, mate, it's just the change of variable.
[math]\int \sin{2x} dx=\frac{1}{2}\int \sin{2x}\ d(2x)=\frac{1}{2}\int \sin{t} dt[/math]

>>8628618
>change of variable
`no its chain rule.
derivate -1/2cos(2x)
you get 2 * 1/2 sin (2x)

derivative of outer times inner function.

>>8628558
You were born.

>>8628695
hey a really helpful person joined why dont you get polite ?