# Beginning calculus trig-related integral

• Jan 17th 2013, 04:56 PM
euphony
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
• Jan 17th 2013, 05:19 PM
Prove It
Re: Beginning calculus trig-related integral
Since you have \displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = i\,y \end{align*}, what does that mean \displaystyle \displaystyle \begin{align*} \frac{dy}{y} \end{align*} is?
• Jan 17th 2013, 05:26 PM
euphony
Re: Beginning calculus trig-related integral
idx.
• Jan 17th 2013, 05:30 PM
Prove It
Re: Beginning calculus trig-related integral
Yes, so now that you have \displaystyle \displaystyle \begin{align*} \frac{dy}{y} = i\,dx \end{align*} you should be able to integrate both sides.
• Jan 17th 2013, 05:41 PM
euphony
Re: Beginning calculus trig-related integral
ln(y) = ∫idx
• Jan 17th 2013, 09:14 PM
Prove It
Re: Beginning calculus trig-related integral
First of all, \displaystyle \displaystyle \begin{align*} \int{\frac{dy}{y}} = \ln{|y|} \end{align*}, not \displaystyle \displaystyle \begin{align*} \ln{(y)} \end{align*}. And what would be \displaystyle \displaystyle \begin{align*} \int{i\,dx} \end{align*}? Remember that \displaystyle \displaystyle \begin{align*} i \end{align*} is a constant.
• Jan 17th 2013, 09:17 PM
euphony
Re: Beginning calculus trig-related integral
ln|y| = ix + c
• Jan 17th 2013, 09:23 PM
euphony
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?
• Jan 18th 2013, 07:49 AM
euphony
Re: Beginning calculus trig-related integral
So if
ln|y| = ix + 0
then would the equivalent form be
y = e^(ix)
• Jan 18th 2013, 03:43 PM
Prove It
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.

\displaystyle \displaystyle \begin{align*} \ln{|y|} &= i\,x + C \\ |y| &= e^{i\,x + C} \\ |y| &= e^C e^{i\,x} \\ y &= A\,e^{i\,x} \end{align*}

Now use your boundary condition to evaluate this constant A.
• Feb 11th 2013, 02:57 PM
euphony
Re: Beginning calculus trig-related integral
y = Ae^(ix)
(1) = A * e^[i(0)]

A = 1?
• Feb 11th 2013, 03:07 PM
Prove It
Re: Beginning calculus trig-related integral
Yes much better :)