Integral involving inverse trig functions. Please help! Final is tomorrow!?

i) Let cos(3x) = t; differentiating, -3sin(3x) d x= dt

==> sin(3x) dx = -dt/3

ii) So integral changes to: (-1/3)∫dt/(16 + t²) = (-1/3)*(1/4)*tan⁻¹(t/4) + C

[Application of ∫dx/(a² + x²) = (1/a)*tan⁻¹(x/a) + C]

Replacing the t from the substitution,

Answer: (-1/12)*tan⁻¹{cos(3x)/4} + C

Integral {sin (3x) dx / (16 + cos²(3x)}

Put cos (3x) = 4 tan α ---------- (1)

Differentiating, - 3sin (3x) dx = 4sec² α dα

=> sin (3x) dx = - (4/3) sec² α dα

=> Numerator = N = - (4/3) sec² α dα -------- (2)

Denominator = D = 16 + 16 tan² α = 16(1 + tan² α) = 16sec² α -------- (3)

Now the integral reduces to :

{-(4/3)/16} * Integral {sec² α dα / sec² α}

= (- 1/12) * Integral dα ---------

= (- 1/12) α + C , where C is an arbitrary constant. ------ (4)

From (1), tan α = cos (3x) / 4

=> α = arc tan {cos(3x) / 4}

Hence, the value of the Integral = (- 1/12) arc tan {(cos3x / 4)} + C

The complicated exponential version of ArcTan(x) is: (a million/2) i (Log(a million - i x) - Log(a million + i x)) The imperative of this is (a million/2) x i (Log(a million - i x) - Log(a million + i x)) - (a million/2) Log(a million + x²) however the 1st 2 words is quite x ArcTan(x), so as that the respond is: x ArcTan(x) - (a million/2) Log(a million + x²)