Quick question - trig integration (1 Viewer)

[karnage]

Hi folks got a question that im stuck on.

Integral of sin.e^2x/e^2x

Edit: I think this is it, please check it math guru's

So i could write it as

Integral of 1/e^2x . sin.e^-2x

= - 1/(e^2x.-2e^-2x) . cos.e^-2x + C

= + 1/2 . cos.e^-2x + C

 

[doink]

sin^(2x) makes no sense and neither does your working,

[sin^(2)x]^-1 =/= sin^(-2)x

is it meant to be sin x e^(2x) ??? even so [sin(e^2x)]^-1 =/= sin(e^-2x)

 

[karnage]

Ah my bad, must have confused you with my typing.

Its the integral of sin(e^-2x) over e^2x

e for exponential

Ive added .'s to ease the confusion.

 

[YannY]

Ah my bad, must have confused you with my typing.

Its the integral of sin(e^-2x) over e^2x

e for exponential

Ive added .'s to ease the confusion.

u=e^-2x
dx=du/[2e^(-2x)]

sub that in you get 1/2sinudu
=-1/2cosu+C
=-1/2cos(e^-2x)

 

[doink]

u=e^-2x then

du/dx = -2e^-2x then

dx = -du/[2e^(-2x)]