Thread: Trying to integrate cos 5x

can some please explain this to me:

if dv = cos 5x dx, how does v = 1/5 sin 5x ??

--i get something totally different (5/2 x^2 sin 5x )

Thanks for any help!

Originally Posted by adam21

can some please explain this to me:

if dv = cos 5x dx, how does v = 1/5 sin 5x ??

--i get something totally different (5/2 x^2 sin 5x )

Thanks for any help!

Here

You can check your answers by differentiating the answer you got when you integrate.

If you differentiate 5/2 x^2 sin 5x you don't get cos(5x), but if you differentiate (1/5)sin(5x) you do.

i'm not sure where you made your mistake, but i think you were mixing up the product rule somewhere in all this, but you don't have a product

Originally Posted by adam21

can some please explain this to me:

if dv = cos 5x dx, how does v = 1/5 sin 5x ??

--i get something totally different (5/2 x^2 sin 5x )

Thanks for any help!

Note that the 5x is the argument of the cosine function; it is not mulitpling anything.

Int[cos(5x)dx]

Let y = 5x --> dy = 5dx

Thus:
Int[cos(5x) dx] = Int[cos(y) * (1/5) dy] = (1/5)*Int[cos(y) dy]

= (1/5)*sin(y) + C = (1/5)*sin(5x) + C

-Dan