TITLE: If $[F(x)]^{100} = \int_{0}^{x} (F(t))^{100} \frac{dt}{1+\sin t}$ then find $F(x)$
QUESTION [5 upvotes]: If $[F(x)]^{100} = \int_{0}^{x} (F(t))^{100} \frac{dt}{1+\sin t}$ then find $F(x)$.

My attempt
Differentiating both sides,
$$100[F(x)]^{99} \frac{d F(x)}{dx} = \frac{F(x)^{100}}{1 + \sin x}$$
then
$$\frac{d F(x)}{F(x)} = \frac{dx}{100(1+\sin x)}$$
and
$$\int \frac{d F(x)}{F(x)} = \int \frac{dx}{100(1+\sin x)}$$
$$\log F(x) = -1/(50+50 \tan (x/2))$$
Hence
$$F(x) = \exp(-1/(50+50\tan (x/2))$$
But, I am not getting my answer right. Where did I go wrong?

REPLY [3 votes]: Note that we have the stationary solution $F(x)\equiv 0$. If $F(x)\not=0$ then, by separation of variables,
$$\int \frac{d F(x)}{F(x)} = \int \frac{dx}{100(1+\sin x)} $$
which implies
$$\log |F(x)| = -\frac{1}{50(1+\tan (x/2))}+c.$$
and for $x\in (-\pi,\pi)$,
$$F(x)=C\exp\left(-\frac{1}{50(1+\tan (x/2))}\right).$$
Moreover by assumption, it seems that $F(0)=0$.