If ∫ cosθ 5 + 7sinθ - 2 ^ 2 θ dθ = Alog _ e | B ( θ ) | + C, where...

$I = ∫ \dfrac{cos\theta}{2\sin^{2}\theta + 7sin\theta + 3}d\theta$

$sin\theta = t ⇒ cos\thetad\theta = dt$

$= \dfrac{1}{2} ∫ \dfrac{1}{{t}^{2} + \dfrac{7}{2}t + \dfrac{3}{2}}dt = \dfrac{1}{2} ∫ \dfrac{1}{{\left(t + \dfrac{7}{4}\right)}^{2} - {\left(\dfrac{5}{4}\right)}^{2}}dt = \dfrac{1}{5}\ln \left|\dfrac{2t + 1}{t + 3}\right| + c = \dfrac{1}{5}\ln \left|\dfrac{2sin\theta + 1}{sin\theta + 3}\right| + c$

so $A = \dfrac{1}{5}$

$B ( \theta ) = \dfrac{5 ( 2sin\theta + 1 )}{sin\theta + 3}$