Hi all!
I'm trying to find the anti derivative of a(t)=cos(t)+sin(t)

so far I have sin(t)+c-cos(t) as the antiderivitive but Im confused because the antiderivitive of cos(t) is sin(t)+c and the antiderivitive of sin(t) is -cos(t)+c
so should the antiderivitive of a(t) be sin(t)+c-cos(t)+D??
just confused about the c/d thing.

Thanks!

The antiderivative of cos(t)+sin(t) would be

sin(t)-cos(t)+C

No need for another constant.

\int \left[cos(t)+sin(t)\right]dt=sin(t)-cos(t)+C

Now, when we take the derivative of sin(t)-cos(t)+C we get back where we started. To cos(t)+sin(t).

Because the derivative of a constant, C, is 0.

Hi all!
I'm trying to find the anti derivitive of a(t)=cos(t)+sin(t)

so far I have sin(t)+c-cos(t) as the antiderivitive but Im confused because the antiderivitive of cos(t) is sin(t)+c and the antiderivitive of sin(t) is -cos(t)+c
so should the antiderivitive of a(t) be sin(t)+c-cos(t)+D ???
just confused about the c/d thing.

Thanks!


It is impossible to identify the constant of integration. The solution which you seek lacks a closed form .

It is impossible to identify the constant of integration. The solution which you seek lacks a closed form .

Huh??
I'm just going to go with +c and be done with it.

Thanks anyway.