$\int{\cos(x)\cosh(x)+\sin(x)\sinh(x)}dx$

Question

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 $$I=sin(x)cosh(x)$$

Answer 1

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 $$\int (\sin (101x) \cdot \sin^{99}x)dx$$

Answer 2

Hint...try differentiating $\sin x\cosh x$

Answer 3

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.

Answer 4

$$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx$$ 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, $$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x))+\frac12(\sin(x)\cosh(x)-\cos(x)\sinh(x))$$ $$=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x)+\sin(x)\cosh(x)-\cos(x)\sinh(x))$$ $$=\frac12(2 \sin(x)cosh(x))$$ $$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx=\sin(x)\cosh(x)+C$$