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Please find this integral1/1+sin2x dx.....................

Ashok Eswar , 7 Years ago
Grade 12
anser 1 Answers
Arun
Dear Ashok
 
1 /1 + sin2x = 1 / sin²x + cos ²x + 2 sinx cosx 
 
 = 1 / (sin x + cos x)²
 
Now divide by cos²x in both numerator and denominator.
 
Hence
 
sec² x/ (1 + tanx)²
Now integrate
 
1 + tan x = t
 
sec² x dx = dt
 
dt/ t² = -1/t =
 = -1 /(1 + tan x) + c
Last Activity: 7 Years ago
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