I have no clue where to start with this any advice would be really appreciated. I tried using u-substitution but that did not work.
Solve the initial value problem
dy/dx = 1-sin(5x)/cos(5x)+5x+6 (x>-1), y=3 when x=0
You need to use paretheses. Is this what you mean?
$\dfrac{dy}{dx} = 1-\dfrac{\sin(5x)}{\cos(5x)}+5x+6$
or did you mean:
$\dfrac{dy}{dx} = \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}$
Assuming it is the latter:
Let $u = \cos(5x)+5x+6$. Then $du = \left(5-5\sin(5x)\right)dx = 5\left(1-\sin(5x)\right)dx$
This gives:
$\displaystyle y = \int \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}dx = \dfrac{1}{5}\int u^{-1}du = \dfrac{1}{5}\ln\left|\cos(5x)+5x+6\right| + C$
Plugging in the initial values:
$3 = \dfrac{1}{5}\ln(7)+C$ shows $C = 3-\dfrac{1}{5}\ln(7)$
This gives $y = \dfrac{1}{5}\ln \left| \dfrac{\cos(5x)+5x+6}{7} \right|+3$