Proving function f(x) is a PDf given integral relationships

[Saracen Rue]

Homework Statement

Prove that $\int _0^a\left(\frac{f\left((sin(ln(c))x\right)+\sqrt{\cos \left(e-\pi ^2\right)}}{\ln \left(\pi ^2-e\right)+\pi ^2\sqrt{\cos \left(e-\pi ^2\right)}}\right)dx$ is a probability density function (when $a=\frac{1}{\pi ^2}$) given that $\int _0^{\pi ^2}f\left(x\right)dx=\ln \left(\pi ^2-e\right)$ and that $c$ is a maximum.

Homework Equations

$\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)$

The Attempt at a Solution

I'm honestly completely stuck with this question. I know that $\int _0^a\left(\frac{f\left((sin(ln(c))x\right)+\sqrt{\cos \left(e-\pi ^2\right)}}{\ln \left(\pi ^2-e\right)+\pi ^2\sqrt{\cos \left(e-\pi ^2\right)}}\right)dx$ can be expressed as $\frac{1}{\ln \left(\pi ^2-e\right)+\pi ^2\sqrt{\cos \left(e-\pi ^2\right)}}\int _0^a\left(f\left((sin(ln(c))x\right)+\sqrt{\cos \left(e-\pi ^2\right)}\right)dx$, but I am unsure of how this helps me to prove that the integral is equal to 1 (thus proving it is a PDf). By specifying that $c$ is a maximum, the question insinuates that $c$ is a variable that can be expressed in terms of another variable which in turn can be derived and solved, however I am unsure of how to form such a relationship.

 

[mfb]

c is a maximum of what?

There are mismatched brackets related to f.
It looks like the problem depends on c, which is odd. What exactly do you know about c?