Question

integration of 1-sinx/1+sinx dx​

Answer 1

Answer:

integral:= 2 tan x- 2 sec x- x + c

Step-by-step explanation:

$\frac{1-sin x}{1+sinx}$ * $\frac{1-sin x}{1-sin x} =\frac{(1-sinx)^{2} }{1-sin^{2} x }$

=$\frac{1-2 sin x + sin^{2} x}{cos^{2}  x} =\frac{1-2 sin x+1- cos^{2} x }{cos^{2} x } \\ \\=\frac{2}{cos^{2} x } - \frac{2sin x}{cos^{2} x } - \frac{cos^{2} x}{cos^{2} x } \\ \\  = 2sec^{2}  x - 2 tan x sec x-1$

∴ $\int {\frac{1- sin x}{1+ sinx} } \, dx =\int {2 sec^{2} x -2 sec x tan x-1 } \, dx \\ \\= 2 tan x-2 sec x-x+c$