In the field of Aerospace, for a thin surface to generate zero vertical force, the inclination it has to be kept at, can be written in the form,
$$
\alpha_0 = \frac{1}{\pi}\int_{0}^{\pi} y_x(\theta)(1-\cos\theta)d\theta
$$
where $y_x(\theta)$ is the slope of the thin surface.
If the thin surface is represented as a discrete Fourier series, the slope comes out to be,
$$
y_x(\theta) = \frac{2}{\sin\theta}\sum_{n=1}^{N}nA_n\cos{k_n\theta} \quad, \quad n \in \mathbb{N}
$$
I'm trying to obtain a closed form solution, for $n$ terms in $y_x$, by solving the resulting integral.
$$
\alpha_0 = \frac{2}{\pi}\sum_{n=1}^{N} nA_n \int_{0}^{\pi} \cos{k_n\theta} \big(\frac{1-\cos\theta}{\sin\theta}\big) d\theta
$$
With some manipulation, this further simplifies to the form,
$$
I = \int_{0}^{\pi/2} \cos{m\phi}\tan{\phi}\ d\phi
$$
where $m \in \mathbb{R}$
Any effort/hint is hugely appreciated!
Best Answer
$m$ must be an odd integer for the singularity at $\phi=\pi/2$ to disappear.
So, assume $m=2n+1$ is positive (as we may); then $$\int_0^{\pi/2}\cos m\phi\tan\phi\,d\phi\underset{\phi=\pi/2-t}{\phantom{\big[}=\phantom{\big]}}(-1)^n\int_0^{\pi/2}\sin(2n+1)t\cot t\,dt=(-1)^n(I_n+I_{n+1}),$$ where $$I_n=\int_0^{\pi/2}\frac{\sin 2nt}{2\sin t}\,dt=\int_0^{\pi/2}\sum_{k=0}^{n-1}\cos(2k+1)t\,dt=\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}.$$