NDA Math Definite Integrals

The above case can be broken down into two sub parts.
Case I:
M is odd.
Therefore the above integral transforms into
I=

π

∫

0

sin‌2‌m‌x

sin‌x

dx
=

π

∫

0

2‌sin‌m‌x.cos‌m‌x

sinx

dx
2I=0
I=0
Case II :
M is even. Hence M-1 will be odd.
Therefore
=2‌

π

∫

0

cos(2m−1)xdx
m∊2n
∴2m−1∊4n−1
∴2m−1 is odd
Im=Im−1
From case I
Im−1=0
Since m ∊ 2n
∴m−1∊2n−1
Therefore
Im=0
Hence the value of the above integral is always 0.