If ∫cosθ5+7sinθ-2cos2θdθ=Aloge|B(θ)|+C, Where C is a constant of integration, then B(θ)A can be

substitute sinθ=t⇒cosθdθ=dt in the given integrand. Then ∫cosθdθ5+7sinθ-2(1-sin2 θ)=∫dt3+7t+2t2=∫dt(2t+1)(t+3)=15∫22t+1-1t+3dt=15logθ2sinθ+1sinθ+3+c Therefore, B(θ)A=5(2sinθ+1)sinθ+3