Question

integration of 1/sin2x

Answer 1

integration of 1/sin2x dx

f(x) dx = cosec 2x dx
          = 1/2 * cosec 2x  d(2x)

∫ f(x) dx = - 1/2 * Ln | Cosec2x + Cot 2x| + K

as we know the standard formula for integration of Cosec x dx

Answer 2

I = ∫dx/sin2x
we know,
sin2x = 2sinx.cosx

I = ∫dx/2sinx.cosx
= 1/2 ∫1.dx/sinx.cosx
now, put 1 = sin²x + cos²x

I =1/2 ∫(sin²x + cos²x)dx/sinx.cosx
=1/2[ ∫tanx.dx + ∫cotx.dx ]
= 1/2[ ln|secx| + ln|sinx|] + C