Consider the integral $$ \int_0^\pi \frac {\cos(\theta)} {f(\sin(\theta))}d\theta $$ Assume that $f(\sin(\theta))$ is nonzero on $[0,\pi]$. Can we use the substitution $u=\sin(\theta)$ to make the integral trivial ($\int_0^0$) and get this is zero for any function $f$?
What about if $f(\sin(\theta))$ is zero within $[0,\pi]$? For example, $f(\sin(\theta))=\sqrt{1-\sin^3\theta}$
Note that before applying integration by substitution to any function $f$ there is some conditions to verify and one of these conditions is that $f$ must be continuously differentiable. so in your case if $f(\sin(\theta_0))=0$ for some $\theta_0\in[0,\pi]$ so you can't apply integration by substitutions.
And to find the value of the integral if it exists you might use other Technics.
No, that's not really what's going on here. The substitution has to be a one-one mapping, or you lose parts of the integral (think of $u=x^2$: then if $x$ ranges between $-1$ and $1$, $u$ goes from $1$ to $0$ and back again).
Instead, look at $$ I = \int_{-\pi/2}^{\pi/2} \frac{\cos{(\phi+\pi/2)}}{f(\sin{(\phi+\pi/2)})} \, d\phi. $$ This is the same function, with $\phi=\theta+\pi/2$. Now, $\cos{(\phi+\pi/2)}=-\sin{\phi}$ and $\sin{(\phi+\pi/2)}=\cos{\phi}$. Hence $$ I = \int_{-\pi/2}^{\pi/2} \frac{-\sin{\phi}}{f(\cos{\phi})} \, d\phi, $$ and this is an odd function, hence the integral is zero.
You can always transform it like this:
$$\int_0^{\pi} \frac{\cos \theta}{f(\sin \theta)} d \theta = \int_0^{\frac{\pi}{2}} \frac{\cos \theta}{f(\sin \theta)} d\theta + \int_{\frac{\pi}{2}}^{\pi} \frac{\cos \theta}{f(\sin \theta)} d\theta = \int_0^{\frac{\pi}{2}} \frac{1}{f(\sin \theta)} d\sin\theta + \int_{\frac{\pi}{2}}^{\pi} \frac{1}{f(\sin \theta)} d\sin\theta = \int_0^{1} \frac{1}{f(u)} du + \int_{1}^{0} \frac{1}{f(u)} du $$
and work only with $f$.
It depends of $f$ if the value exists. If $\int_0^{1} \frac{1}{f(u)} du \in \mathbb{R}$, then of course the sum will be $0$, otherwise nothing more could be said.
