Why isn't $\int\sin(ix)~dx$ equal to $i\cos(ix)+C$ ?

Question

I was playing around with imaginary numbers, and I tried to solve $$\int\sin(ix)~dx$$ and ended up getting $$i\cos(ix)+C$$

But apparently the answer is $$i\cosh(x)+C$$

I was just wondering, is this correct? And what does the "$h$" stand for/mean? Where did it even come from. Thanks in advance.

Answer 1

Here are two useful definitions / relations

$$\cosh(x) = \frac{e^x + e^{-x}}{2}$$

$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$

Using these definitions you can see that

$$\cos(ix) = \frac{e^{i(ix)} + e^{-i(ix)}}{2} = \frac{e^{-x} + e^x}{2} = \cosh (x)$$

So you did get the same answer, but you just had it in a different form.

Answer 2

Using $$ \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\quad\text{and}\quad\cos(x)=\frac{e^{ix}+e^{-ix}}{2} $$ we have $$ \cos(ix)=\cosh(x)\quad\text{and}\quad\sin(ix)=i\sinh(x) $$ Thus, $$ \begin{align} \sin(x+iy) &=\sin(x)\cos(iy)+\cos(x)\sin(iy)\\ &=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\\ \end{align} $$ and $$ \begin{align} \cos(x+iy) &=\cos(x)\cos(iy)-\sin(x)\sin(iy)\\ &=\cos(x)\cosh(y)-i\sin(x)\sinh(y)\\ \end{align} $$