If ∫ cos θ 5 + 7 sin θ - 2 cos 2 θ d θ = Alog e | B ( θ ) | + C , where C is a constant of integration, then B ( θ ) A can be:

Step-by-step solution

I = ∫ cos θ 2 sin 2 θ + 7 sin θ + 3 d θ sin θ = t    ⇒    cos θ d θ = d t = 1 2 ∫ 1 t 2 + 7 2 t + 3 2 d t = 1 2 ∫ 1 t + 7 4 2 - 5 4 2 d t = 1 5 ln 2 t + 1 t + 3 + c = 1 5 ln 2 sin θ + 1 sin θ + 3 + c so A = 1 5 B ( θ ) = 5 ( 2 sin θ + 1 ) sin θ + 3