How to solve this integral

dx/(1-sinx)

multiply top and bottom by (1 + sin(x)

∫ (1 + sin(x)) / (1 - sin^2(x)) dx
= ∫ (1 + sin(x)) / (cos^2(x)) dx
= ∫ sec^2(x) + sin(x)/cos^2(x) dx

∫(sec^2(x) + sin(x)/cos^2(x) dx)
= ∫(sec^2(x) dx) + ∫(sin(x)/cos^2(x) dx)
let u = cos(x) so du = -sin(x)
= ∫(sec^2(x) dx) - ∫(1/(u^2) dx)
= tan(x) + 1/u + C
= tan(x) + 1/cos(x) + C
= tan(x) + sec(x) + C
C is a constant

... ∫ [ 1 / ( 1 - sin x ) ] dx

= ∫ { 1 / [ 1 - cos ( π/2 - x ) ] } dx

= ∫ { 1 / [ 2. sin² ( π/4 - x/2) ] } dx

= ( 1/2 )· ∫ csc² ( π/4 - x/2 ) dx

= ( 1/2 )· [ - cot ( π/4 - x/2 ) ]· [ 1/ (-1/2) ] + C

= cot [ (π/4) - (x/2) ] + C ........................................… Ans.
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1/(1-sinx) = (1+sinx)/[(1-sinx)(1+sinx)]
= (1+sinx)/[1-sin^2(x)]
= (1+sinx)/cos^2(x)
= 1/cos^2(x) + sinx/cos^2(x)
= sec^2(x) + (sinx/cosx)(1/cosx)
= sec^2(x) + tanxsecx

so
int [1/(1-sinx)]dx = int [sec^2(x) + tanxsecx ] dx
= tanx + secx + C

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