# Thread: Beginning calculus trig-related integral

1. ## 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.

2. ## 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?

idx.

4. ## 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.

ln(y) = ∫idx

6. ## 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.

7. ## Re: Beginning calculus trig-related integral

ln|y| = ix + c

8. ## 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?

9. ## Re: Beginning calculus trig-related integral

So if
ln|y| = ix + 0
then would the equivalent form be
y = e^(ix)

10. ## Re: Beginning calculus trig-related integral

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.

11. ## Re: Beginning calculus trig-related integral

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

A = 1?

12. ## Re: Beginning calculus trig-related integral

Yes much better