If $\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^{2} \theta} d \theta=A \log _{e}|B(\theta)|+C$, where $C$ is a constant of integration, then $\frac{\mathrm{B}(\theta)}{\mathrm{A}}$ can be:

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

(A) $\frac{5(\sin \theta+3)}{2 \sin \theta+1}$

(C) $\frac{2 \sin \theta+1}{5(\sin \theta+3)}$

(D) $\frac{2 \sin \theta+1}{\sin \theta+3}$

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Indefinite Integrals

\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:

Answer: (B)

Writing

cos^{2} θ = 1 - sin^{2} θ

, the denominator becomes

5 + 7 sin θ - 2 (1 - sin^{2} θ) = 2 sin^{2} θ + 7 sin θ + 3

. Substituting

t = sin θ

d t = cos θ d θ

, the integral is

\int \frac{d t}{2 t ^{2} + 7 t + 3} = \int \frac{d t}{( 2 t + 1 ) ( t + 3 )}

. Partial fractions give

\frac{1}{( 2 t + 1 ) ( t + 3 )} = \frac{2}{5 ( 2 t + 1 )} - \frac{1}{5 ( t + 3 )}

, and integrating yields

\frac{1}{5} ln \frac{2 t + 1}{t + 3}

. Thus

A = \frac{1}{5}

and

B (θ) = \frac{2 s i n θ + 1}{s i n θ + 3}

, so

\frac{B ( θ )}{A} = \frac{5 ( 2 s i n θ + 1 )}{s i n θ + 3}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option