Why am I getting the wrong answer? Calculus, trig substitution.

[MonkeySage]

My problem is Int(sqrt(16-x²)/x)

so u=x, a=4, and x=4sinθ

dx=4cosθdθ and sqrt(16-x²)=4cosθ

So what I get is int(cotθdθ)= ln(abs(sinθ))+C or ln(abs(x/4))+C

But this is apparently not the correct answer. Somehow the correct answer is 4ln(abs(4-sqrt(16-x²)))/x + sqrt(16-x²) + C

And yes, my instructions specifically ask me use the substitution x=4sinθ

[jayem]

What are you integrating with respect too in the first part?
(though in all honesty I can't think how to integrate that simply, and I'm too tired to try.)
Though I think we using the sin^2+cos^2=1 back (or possibly not using that in the first place) gets two separate single terms.

[crayzz]

My problem is Int(sqrt(16-x²)/x)

so u=x, a=4, and x=4sinθ

dx=4cosθdθ and sqrt(16-x²)=4cosθ

So what I get is int(cotθdθ)= ln(abs(sinθ))+C or ln(abs(x/4))+C

But this is apparently not the correct answer. Somehow the correct answer is 4ln(abs(4-sqrt(16-x²)))/x + sqrt(16-x²) + C

And yes, my instructions specifically ask me use the substitution x=4sinθ

After substitution, shouldn't you be getting int 4cot(θ)cos(θ)? I think you forgot to include the factor of 4cos(θ) that comes from subtituting x = 4sin(θ) into dx.

[jayem]

After substitution, shouldn't you be getting int 4cot(θ)cos(θ)? I think you forgot to include the factor of 4cos(θ) that comes from subtituting x = 4sin(θ) into dx.

That was my guess too, though not very well explained (I didn't want to help too much too quickly).
...
Then by [using an identity you used before*] you are then left to integrate a sum of two trig terms (one I'd have to look up and one that is easy).

*It does need doing twice.

[Kato]

So, before I make a lengthy try to type formulas into thread messages... does the problem still exist or has it been answered? Because I'd rather not have to do that if it's not needed.