Using substitution to make an integral trivial 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, $f(\sin(\theta))=\sqrt{1-\sin^3\theta}$
 A: 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.
A: 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.
A: 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.
