Is reduction formula the only way?

Let ${ I }_{ m }=\int _{ 0 }^{ \pi }{ \frac { 1-\cos { mx } }{ 1-cosx } dx }$

Now, for m=1,2,3 , ${ I }_{ m }$ = $\pi ,2\pi ,3\pi$ . So, claim : ${ I }_{ m }$ = $m\pi$

Proof by induction:

As ${ I }_{ 1 }$ is true , let the result be true for all positive integers $\le k$ where $k\in N$

Then, ${ I }_{ k }=\int _{ 0 }^{ \pi }{ \frac { 1-\cos { kx } }{ 1-cosx } dx } =k\pi ,\quad for\quad k=1,2,...,k-1,k\quad$ ${ I }_{ k+1 }-{ I }_{ k }=\int _{ 0 }^{ \pi }{ \frac { (1-\cos { (k+1)x } )-(1-coskx) }{ 1-cosx } dx }$ $=\int _{ 0 }^{ \pi }{ \frac { coskx-cos(k+1)x }{ 1-cosx } dx }$ $=\int _{ 0 }^{ \pi }{ \frac { 2sin\left( k+\frac { 1 }{ 2 } \right) x.sin\left( \frac { x }{ 2 } \right) }{ 2{ sin }^{ 2 }\left( \frac { x }{ 2 } \right) } dx }$ $=\int _{ 0 }^{ \pi }{ \frac { sin\left( k+\frac { 1 }{ 2 } \right) x }{ { sin }\left( \frac { x }{ 2 } \right) } dx }$ Now substituting k = k-1 ${ I }_{ k }-{ I }_{ k-1 }=\int _{ 0 }^{ \pi }{ \frac { sin\left( k-\frac { 1 }{ 2 } \right) x }{ sin\left( \frac { x }{ 2 } \right) } dx }$ $\therefore ({ I }_{ k+1 }-{ I }_{ k })-({ I }_{ k }-{ I }_{ k-1 })=\int _{ 0 }^{ \pi }{ \frac { sin\left( k+\frac { 1 }{ 2 } \right) x-sin\left( k-\frac { 1 }{ 2 } \right) x }{ sin\frac { x }{ 2 } } dx }$ $=\int _{ 0 }^{ \pi }{ \frac { 2coskx\quad sin\frac { x }{ 2 } }{ sin\frac { x }{ 2 } } dx }$ $={ 2\left[ \frac { sinkx }{ k } \right] }_{ 0 }^{ \pi }=0$ So, ${ I }_{ k+1 }=2{ I }_{ k }-{ I }_{ k-1 }$

${ I }_{ k+1 }=2k\pi -(k-1)\pi$

Hence, ${ I }_{ k+1 }=(k+1)\pi$ So, by induction ${ I }_{ m }=m\pi \quad \forall \quad m\in N$

${ I }_{ 74 }=74\pi =\frac { 444 }{ 6 } { \pi }^{ 1 }$ a+b=445