Calculate the integral of y=sin6x/(5+cos12x).

We'll write the term from denominator cos 12x = cos
2*(6x)

We'll apply the
formula:

cos 2a = 2(cos a)^2 -
1

cos 2*(6x) = 2(cos 6x)^2 -
1

The denominator will
become:

5+cos12x = 5 + 2(cos 6x)^2 -
1

5+cos12x = 2(cos 6x)^2 +
4

2(cos 6x)^2 = 2[(cos 6x)^2 +
2]

Int f(x)dx = (1/2)Int sin 6xdx/[(cos 6x)^2 +
2]

We'll substitute cos 6x  =
t.

We'll differentiate both
sides:

- 6 sin 6x dx = dt

sin
6x dx = -dt/6

We'll write the integral in
t:

Int f(x)dx = (-1/12)Int dt/(t^2 +
2) 

 (-1/12)Int dt/(t^2 + 2)  = -(1/12*sqrt2)*arctan
t/sqrt2 + C

Int f(x)dx = -(1/12*sqrt2)*arctan (cos
6x)/sqrt2 + C