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प्रश्न
Find : ` int (sin 2x ) /((sin^2 x + 1) ( sin^2 x + 3 ) ) dx`
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उत्तर १
` 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 `
उत्तर २
`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`
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