Verified answer

∫(cosx/(3sinx+12))(dx)

= (1/3) ∫(cosx/(sinx+4))(dx) (which makes the numerator the derivative of the denominator)

= (1/3) ln |sinx + 4| + C

About the people that are saying:

(1/3) ln |3sinx + 12| + C

= (1/3) ln |3(sinx + 4)| + c

= (1/3) (ln3 + ln |sinx + 4| ) + C

= (1/3) ln3 + (1/3) ln |sinx + 4| + C

= (1/3) ln |sinx + 4| + K

It's the same answer, just a different constant

K = C + ln(3)/3

Let u = 3sinx + 12

du = 3cosx dx

dx = du/3

Integral 1/u du/3 = 1/3 ln u = 1/3 ln|3sinx + 12| + C

u = 3*sin(x) + 12

du = 3cos(x), (1/3)du = cos(x)

1/3*integral( du/u) = (1/3)ln|(u)| =

(1/3)*ln|3sin(x) + 12| + C