Find the positive value of $k$ for which the value of the definite integral $\int_0^{\frac{\pi}2}|\cos x-kx|dx$ is minimized

Question

Find the positive value of $k$ for which the value of the definite integral $\int_0^{\frac{\pi}2}|\cos x-kx|dx$ is minimized.

If I draw the graph of $\cos x$ and the straight line $kx$, they intersect between $0$ to $\frac{\pi}2$. Let the abscissa of that point of intersection be $\alpha$. So, the given integral can be written as $$\int_0^{\alpha}(\cos x-kx)dx+\int_{\alpha}^{\frac{\pi}2}(kx-\cos x)dx$$

I tried to take its derivative but got $\cos \alpha=k\alpha$, which is the same as the point of intersection. So, don't know how to find $\alpha$ or $k$.

A similar question exists here.

Answer 1

If you calculate that integral you get

$$I = 2\sin (\alpha) -k\alpha^2 +{k\pi^2\over 8}-1$$

Using $k = {\cos (\alpha) \over \alpha}$ you get

$$I(\alpha) = 2\sin (\alpha) -\alpha\cdot \cos (\alpha) +{\cos (\alpha)\over \alpha}{\pi^2\over 8}-1$$ where $\alpha \in [0,{\pi\over 2}]$. Now take the derivative of $I$...