Minimize The Integral

$I\quad =\quad \int _{ 0 }^{ { \pi }/{ 2 } }{ \left| \cos { x } -kx \right| } dx$ .

Diagram Let point $P(\alpha ,\cos { \alpha } )$ on curve $y=cos(x)$

So

$\\ \quad \quad k\alpha \quad =\quad \cos { \alpha } \\ \quad \boxed { k\quad \quad =\quad \quad \cfrac { \cos { \alpha } }{ \alpha } }$ .

$I\quad =\quad \int _{ 0 }^{ \alpha }{ (\cos { x } -kx) } dx\quad +\quad \int _{ \alpha }^{ { \pi }/{ 2 } }{ (kx } -\cos { x) } dx$ .

Now after integrating it we get function of $I$ in term's of $\alpha$

$I(\alpha )\quad =\quad 2\sin { \alpha } \quad -\quad \cfrac { \alpha \cos { \alpha } }{ 2 } \quad +\quad \cfrac { \cos { \alpha } }{ 2\alpha } (\cfrac { { \pi }^{ 2 } }{ 4 } \quad -\quad { \alpha }^{ 2 })\quad -1$ .

$\cfrac { d[I(\alpha )] }{ d\alpha } \quad =\quad 0$ .

$\cfrac { d[I(\alpha )] }{ d\alpha } \quad =\quad 0\\ \\ \tan { (\alpha ) } =\quad \cfrac { -1 }{ \alpha } \quad \quad \times \quad (\quad rejected\quad \because \quad \alpha \quad >\quad 0\quad \quad \& \quad \tan { (\alpha ) } >\quad 0\quad )\\ \\ \quad or\quad \\ \\ \boxed { \alpha \quad =\quad \quad \cfrac { \pi }{ \sqrt { 8 } } }$ .

$\Rightarrow \quad k\quad =\quad \cfrac { \sqrt { 8 } }{ \pi } (\cos { (\cfrac { \pi }{ \sqrt { 8 } } ) } )\\ \Rightarrow \quad a\quad =\quad 8\\ \Rightarrow { \quad a }^{ 2 }\quad =\quad 64\quad \\ \\$ .

Q.E.D

Consider the equation, $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left| cosx\quad -\quad kx \right| \quad dx }$

Let us assume it is negative from 0 to some h and then becomes positive from h to $\frac { \pi }{ 2 }$

Then it can be split as

$\int _{ 0 }^{ h }{ (kx-cosx)\quad dx } +\int _{ h }^{ \frac { \pi }{ 2 } }{ (cosx-kx)dx } \\ \\ \\$

Thus effectively removing modulus sign,

now differentiating equation with respect to k and equating it to 0, (for extrema)

(to determine whether its maxima or minima, check yourself)

We get

$\int _{ 0 }^{ h }{ (x)\quad dx } +\int _{ h }^{ \frac { \pi }{ 2 } }{ (-x)dx }$

which gives

$h=\frac { \pi }{ \sqrt { 8 } } \\$

Now at x=h, cosx-kx=0

so

$k=\frac { cos(h) }{ h } =\frac { \sqrt { 8 } }{ \pi } cos(\frac { \pi }{ \sqrt { 8 } } )\\ \\$

which a=8 and its square as 64