Answer 1

` I =  int  (sin 2x ) /((sin^2 x + 1) ( sin^2 x + 3 ) ) dx`

` int  ( 2 sin x cos x  ) /((sin^2 x + 1) ( sin^2 x + 3 ) ) dx`

put t = sin² x

`dt/dx = 2 ` sin x cos x 

⇒ dt = 2 sin x cos dx 

⇒ I = `int  dt/ (( t + 1 )( t + 3) ) dt `

` I = 1/2 [ int 1/(t + 1) dt - int 1/(( t + 3 )) dt ] `

` I = 1/2 [  "In" ( 1+ t) - "In" (3+t)] + C` 

`I = 1/2 "In" ((1+t)/(3 +t)) + C`

`⇒ I = 1/2 " In" ((1 + sin^2 x ) /( 3 + sin^2 x ) ) +  c `

Answer 2

`int_  (sin2"x")/((sin^2 "x"+1)(sin^2"x"+3))d"x"`

⇒ `I  = int_  (2sin"x"·cos"x")/((sin^2 "x"+1)(sin^2"x"+3))d"x"`

let sin² x + 3 = t ⇒ 2sin x·cos xdx = dt

Therefore,

`I = int_  (d"t")/(("t" - 2)"t")`

⇒ `I = 1/2 int_  ((1)/("t"-2)- 1/"t")d"t"`

⇒ `I = 1/2 [ "In" ( "t" -2) - "In"  "t"] + c`

⇒ `I = 1/2 "In" (("t"-2)/("t")) + c`

⇒ `I =  "In" sqrt(("t"-2)/("t")) + c`

⇒ `I = "In" sqrt((sin^2 "x" +1)/(sin^2 "x"+3)) + c`