Let $f$ be a differentiable function satisfying $f(x+y)=(f(x))^{\cos y}\cdot (f(y))^{\cos x}$ for all $x,y \in R$. $f(0)=1, f'(0)=\ln 2$.
If $$\int_{\frac{\pi}{2}}^{\frac{17\pi}{2}} \frac{f(x)}{f(x)+f(\pi /2-x)} dx = k\pi$$ Find the value of $k$. I have tried approaches by substituting $y$ with $\pi /2-x$ in the original equation, but that did not help too much. Taking $\ln$ on both sides of the functional equation and differentiating it is something I tried only because of the fact that $f'(0)$ is given. Trying out properties of definite integrals directly isn not helping either. What am I missing?
Given values: $f(0)=1, f(e)=e$
Given condition: $f(x+y)=(f(x))\cos y⋅(f(y))\cos x f(x+y)=(f(x))\cosy⋅(f(y))\cosx$
Partially differentiating the above with respect to $x$: $f'(x+y)=\cos y[f(x)]^{\cos y-1}\cdot f'(x)\cdot [f(y)]^{\cos x} + [f(x)]^{cosy}\cdot [f(y)]^{\cos x}\cdot \ln [f(y)] \cdot (-\sin x)$
Let $x=0$, therefore, $f'(y)=\cos y \cdot \ln 2 \cdot f(y)$
$\frac{f'(y)}{f(y)}=\cos y \cdot \ln 2$
Integrating both sides:
$\ln f(y)=\ln2 \cdot \sin y + c$
$f(y) = \lambda\cdot 2^{\sin y}$
When $y=0, f(y)=0$, hence, $\lambda=1$.
The required integral is not a particularly difficult one and yields the value $4\pi$. Hence, $k=4$.