Answer 1

`int_  (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`

Put 2 + sin x = t

⇒ 1 + sin x = t - 1

cos x dx = dt

`int_  ("dt")/(("t" -1) "t")`

 = `int_  ((1)/("t" - 1) - (1)/("t")) "dt"`

= `int_  (1)/("t" -1) "dt" - int 1/"t" "dt"`

= log (t - 1) - log t + C

= log (2 + sin x - 1) - log (2 + sin x) + C

= log (1 + sin x) - log (2 + sin x) + C

= `"log" ((1+ sin "x")/(2 + sin "x")) + "C" `   ...`(∵ "log m" - "log n" = "log" ("m"/"n"))`