Question

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

Answer 1

`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`