Question

If I = `int (sin2x)/(3x + 4cosx)^3 "d"x`, then I is equal to ______.

Options

• `(3cosx + 8)/(3 + 4cosx)^2 + "c"`

• `(3 + 8cosx)/(16(3 + 4cos x)^2) + "c"`

• `(3 + cosx)/(3 + 4 cosx)^2 + "c"`

• `(3 - 8cosx)/(16(3 + 4cosx)^2) + "c"`

Answer 1

If I = `int (sin2x)/(3x + 4cosx)^3 "d"x`, then I is equal to `(3 + 8cosx)/(16(3 + 4cos x)^2) + "c"`.

Explanation:

I = `int (sin2x)/(3x + 4cosx)^3 "d"x`

⇒ I = `int (2sinx cosx)/(3 + 4cosx)^3 "d"x`

Put 3 + 4 cos x = t

⇒ cos x = `("t" - 3)/4`

⇒ sin x dx = `"dt"/((-4))`

∴ I = `int (2("dt"/(-4))*(("t" - 3)/4))/("t")^3`

= `(-1)/8 int ("t" - 3)/"t"^3 "dt"`

= `(-1)/8 (int "dt"/"t"^2 - 3 int "dt"/"t"^3)`

= `(-1)/8 ((-1)/"t" + 3/(2"t"^2)) + "c"`

= `(1/(8"t") - 3/(16"t"^2)) + "c"`

= `(2"t" - 3)/(16"t"^2) + "c"`

= `(2(3 + 4cosx) - 3)/(16(3 + 4cosx)^2) + "c"`

∴ I = `(3 + 8cosx)/(16(3 + 4cosx)^2) + "c"`