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
$$I=sin(x)cosh(x)$$
Best Answer
Hint...try differentiating $\sin x\cosh x$