Suppose I have a continuous function $f$ that is always positive over the range of integration and its derivative $f'$. Is there a way to find the following indefinite integral?
$$\int \frac{f'(x)}{f(x)^2}\,\sin(x) \, \mathrm d x$$
Could integration by parts or some clever substitution be used here?
On
Well, you can also use integration by parts to conclude that
$$\int_a^b \frac{f'(x)}{f(x)^2}\,\sin(x) \, \mathrm d x=\int_a^b \frac{1}{f(x)}\,\cos(x) \, \mathrm d x - \left[\frac{\sin(x)}{f(x)}\right]_a^b,$$
so the problem is simplified if $\int\frac{\cos(x)}{f(x)}\,\mathrm dx$ is easier to calculate.
Let $g(x)=\frac{1}{f(x)}$ and let us assume that $f(x)$ is positive and differentiable.
The given integral equals $$\begin{eqnarray*} \int -g'(x)\sin(x)\,dx &\stackrel{\text{IBP}}{=}&C+ g'(x)\cos(x)-\int g''(x)\cos(x)\,dx\\&\stackrel{\text{IBP}}{=}&C+g'(x)\cos(x)-g''(x)\sin(x)+\int g'''(x)\sin(x) \end{eqnarray*}$$ or $$ C+g'(x)\cos(x)-g''(x)\sin(x)+g'''(x)\cos(x)-g^{(4)}(x)\sin(x)+\ldots $$ provided that this expression actually makes sense.