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Evaluate : ∫dx/(1 + sin x), x ∈ [0,π/2]

Answer»

∫dx/(1 + sin x), x ∈ [0,π/2]

Let I = ∫dx/(1 + sin x), x ∈ [0,π/2]

∫(1 - sin x)dx/(1 + sin x)(1 - cos x), x ∈ [0,π/2]

∫(1 - sin x)dx/(1 - sin2 x), x ∈ [0,π/2]

∫(1 - sin x)dx/(cos2 x), x ∈ [0,π/2]

∫((1/cos2 x) - (sin x/cos2 x))dx, x ∈ [0,π/2]

∫(sec2 x - tan x sec x)dx, x ∈ [0,π/2]

∫sec2 x dx, x ∈ [0,π/2] - ∫tan x.sec x dx, x ∈ [0,π/2]

[tan x], x ∈ [0,π/2] - [sec x], ∈ [0,π/2]

((tan π/2) - tan 0) - ((sec π/2) - sec 0)

0 - (-1) 

= 1



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