Beginning calculus trig-related integral

So it starts like this:

**Find the derivative of y = isinx + cosx**

And I don't know what I'm supposed to do with the complex number so I just sort of left it and got this:

dy/dx = -sinx + icosx

Which is equal to i * y

Then it says to

1) find dy/y

2) use the condition x=0 to determine the value of *c* (constant of integration)

3) write this antiderivative in its equivalent form

I don't know what to do next! Please dumb it down as much as possible. :D

Re: Beginning calculus trig-related integral

Since you have , what does that mean is?

Re: Beginning calculus trig-related integral

Re: Beginning calculus trig-related integral

Yes, so now that you have you should be able to integrate both sides.

Re: Beginning calculus trig-related integral

Re: Beginning calculus trig-related integral

First of all, , not . And what would be ? Remember that is a constant.

Re: Beginning calculus trig-related integral

Re: Beginning calculus trig-related integral

So boundary condition x=0

y = isin(0) + cos(0)

y = 1

ln|1| = i(0) + c

c = 0

Is this right? What does it mean to write the antiderivative in its equivalent form?

Re: Beginning calculus trig-related integral

So if

ln|y| = ix + 0

then would the equivalent form be

y = e^(ix)

Re: Beginning calculus trig-related integral

Quote:

Originally Posted by

**euphony** So if

ln|y| = ix + 0

then would the equivalent form be

y = e^(ix)

That is the correct equivalent form, however, your elimination of the constant is incorrect due to the absolute value around the y.

Now use your boundary condition to evaluate this constant A.

Re: Beginning calculus trig-related integral

y = Ae^(ix)

(1) = A * e^[i(0)]

A = 1?

Re: Beginning calculus trig-related integral