Question

Evaluate:

`int (1 + sin "x")/(1 - sin "x") "dx"`

Answer 1

Let I = `int (1 + sin "x")/(1 - sin "x") "dx"`

= `int (1 + sin "x")/(1 - sin "x") xx (1 + sin "x")/(1 + sin "x") "dx"`

= `int (1 + sin "x")^2/(1 - sin^2 "x") "dx"`

= `int (1 + sin^2 "x" + 2 sin "x")/(cos^2 "x")"dx"`

= `int (1/cos^2"x" + sin^2"x"/cos^2"x" + 2 sin "x"/cos^2"x") "dx"`

⇒ I = `sec^2 "x. dx" + int tan^2 "x dx" + 2 int sin "x"/cos^2"x" "dx"`

= `int sec^2 "x dx" + int (sec^2 "x" - 1)"dx" + 2 int sin "x"/cos^2"x" "dx"`

Let cos x = t,

∴ − sin x dx = dt

= `2 int sec^2 "x dx" - int 1 . "dx" - 2 int 1/"t"^2 "dt"`

= `2 tan "x" - "x" + 2/"t" + "c"`

∴ I = `2 tan"x" + 2/cos "x" - "x" + "c"`