integration

sin2 nx
------------------ dx
sin2 x

4 Answers

4
UTTARA ·

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!!)

1
RAY ·

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

1
RAY ·

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

4
UTTARA ·

Welcome :)

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