$\int{\cos(x)\cosh(x)+\sin(x)\sinh(x)}dx$ How would I evaluate the integral:
$$I=\int{(\cos(x)\cosh(x)+\sin(x)\sinh(x)})\,dx$$
My thought was to use:
$$\cos(ix)=\cosh(x)$$
and
$$\sin(ix)=i\sinh(x)$$
or expand all four trig functions into exponentials but this was very messy
EDIT:
If I split this into two integrals where $I=I_1+I_2$
$$I_1=\int{cos(x)cosh(x)}dx$$
$$I_2=\int{sin(x)sinh(x)}dx$$
$$I_1=sin(x)cosh(x)-\int{sin(x)sinh(x)}dx$$
This second part is equal to $I_2$ so does that mean that 
$$I=sin(x)cosh(x)$$
 A: Hint...try differentiating $\sin x\cosh x$
A: Your integrand is not exactly in a suitable form. You'd better consider the following two functions:
$$\cos x\cosh x-i\sin x\sinh x=\cos(x+ix)=\cos(1+i)x,$$
and
$$\cos x\cosh x+i\sin x\sinh x=\cos(x-ix)=\cos(1-i)x.$$
The antiderivatives are simply
$$\frac{\sin(1+i)x}{1+i},\frac{\sin(1-i)x}{1-i}$$ which you can develop in terms of the trigonometric and hyperbolic functions. Finally, take the sum of the real and imaginary parts. And laugh.
A: I think you might want to notice that we have $\sin x$ on right side while it's derivative i.e. $\cos x$ on left side. Also we have $\cosh x$ on left side while it's derivative i.e. $\sinh x$ on the right side.
Don't you think this seems to be much similar to what happens in the product rule. 
Let $f(x)=\sin x$ and $g(x)=\cosh x$ 
Hence the integral can be written as $$\int (f'g +g'f) dx$$
Which simply equals $$f(x)\cdot g(x)+ C$$ by noticing the product rule. 
Hence the answer to integral is $$\sin x\cosh x +C$$
You might also want to solve One such question from MIT Integration bee using similar idea which is $$\int (\sin (101x) \cdot \sin^{99}x)dx$$
A: $$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx$$
First divide this into two integrals
$$\int \cos(x) \cosh(x)dx+\int \sin(x) \sinh(x) dx$$
Then solve for $\int \cos(x) \cosh(x)dx$ using integration by parts and we get,
$$\int \cos(x) \cosh(x)dx=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x))$$
and
$$\int \sin(x) \sinh(x)dx=\frac12(\sin(x)\cosh(x)-\cos(x)\sinh(x))$$
Now add them both together and we get,
$$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x))+\frac12(\sin(x)\cosh(x)-\cos(x)\sinh(x))$$
$$=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x)+\sin(x)\cosh(x)-\cos(x)\sinh(x))$$
$$=\frac12(2 \sin(x)cosh(x))$$
$$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx=\sin(x)\cosh(x)+C$$
