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

Question

If ∫ ( cos θ/5+7 sin θ-2 cos 2 θ) d θ= A log e | B (θ)|+ C where C is a constant of integration, then ( B (θ)/ A ) can be (A) (2 sin θ+1/5( sin θ+3)) (B) (2 sin θ+1/ sin θ+3) (C) (5( sin θ+3)/2 sin θ+1) (D) (5(2 sin θ+1)/ sin θ+3)

Tardigrade Admin · Accepted Answer

∫ ( cos θ d θ/5+7 sin θ-2 cos 2 θ) ∫ ( cos θ d θ/3+7 sin θ+2 sin 2 θ) [sin θ= t , cos θ d θ= dt ] ∫ ( dt /2 t 2+7 t +3)=∫ ( dt /(2 t +1)( t +3)) =(1/5) ∫((2/2 t +1)-(1/ t +3)) dt =(1/5) ln |(2 t +1/ t +3)|+ C =(1/5) ln |(2 sin θ+1/ sin θ+3)|+ C A =(1/5) and B (θ)=(2 sin θ+1/ sin θ+3)

Q. If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be

$\frac{2 s i n θ + 1}{5 ( s i n θ + 3 )}$

$\frac{2 s i n θ + 1}{s i n θ + 3}$

$\frac{5 ( s i n θ + 3 )}{2 s i n θ + 1}$

$\frac{5 ( 2 s i n θ + 1 )}{s i n θ + 3}$

Q. If ∫5+7sinθ−2cos2θcosθ​dθ=Aloge​∣B(θ)∣+C where C is a constant of integration, then AB(θ)​ can be

5(sinθ+3)2sinθ+1​

sinθ+32sinθ+1​

2sinθ+15(sinθ+3)​

sinθ+35(2sinθ+1)​

∫5+7sinθ−2cos2θcosθdθ​ ∫3+7sinθ+2sin2θcosθdθ​[sinθ=t,cosθdθ=dt] ∫2t2+7t+3dt​=∫(2t+1)(t+3)dt​ =51​∫(2t+12​−t+31​)dt =51​ln∣∣​t+32t+1​∣∣​+C =51​ln∣∣​sinθ+32sinθ+1​∣∣​+C A=51​ and B(θ)=sinθ+32sinθ+1​

Q. If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be

$\frac{2 s i n θ + 1}{5 ( s i n θ + 3 )}$

$\frac{2 s i n θ + 1}{s i n θ + 3}$

$\frac{5 ( s i n θ + 3 )}{2 s i n θ + 1}$

$\frac{5 ( 2 s i n θ + 1 )}{s i n θ + 3}$