integration

Let n = p (for convenience)

I = ∫2 (sin² px) / 2(sin²x) dx

= 1/2 ∫(1-cos px)/(sin²x) dx

= 1/2 [ - cot x - ∫ cos px/(sin²x) dx ]

∫ cos px/(sin²x) dx = -cotx cospx + p ∫ cotx sinpx dx

∫ cotx sinpx dx = - cos px (1/p) cotx - ∫1/p cos px cosec²x dx

NOW

∫ cos px/(sin²x) dx = -cotx cospx + p [- cos px (1/p) cotx - ∫1/p cos px cosec²x dx ]

=> - ∫ cos px/(sin²x) dx ] = 0

=> I = -1/2 cot x

( A bit confusing & may have mistakes but i hope itz correct!!)

the limit is 0 to Î /2 and answer is nÎ

thank u for the method uttara....:)