Question

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 $\frac{ B (\theta)}{ A }$ can be

• A. $\frac{2 \sin \theta+1}{5(\sin \theta+3)}$
• B. $\frac{2 \sin \theta+1}{\sin \theta+3}$
• C. $\frac{5(\sin \theta+3)}{2 \sin \theta+1}$
• D. $\frac{5(2 \sin \theta+1)}{\sin \theta+3}$

Answer 1

Answer: (D)

$\int \frac{\cos \theta d \theta}{5+7 \sin \theta-2 \cos ^{2} \theta}$
$\int \frac{\cos \theta d \theta}{3+7 \sin \theta+2 \sin ^{2} \theta} [sin \theta= t , \cos \theta d \theta= dt ]$
$\int \frac{ dt }{2 t ^{2}+7 t +3}=\int \frac{ dt }{(2 t +1)( t +3)}$
$=\frac{1}{5} \int\left(\frac{2}{2 t +1}-\frac{1}{ t +3}\right) dt$
$=\frac{1}{5} \ln \left|\frac{2 t +1}{ t +3}\right|+ C$
$=\frac{1}{5} \ln \left|\frac{2 \sin \theta+1}{\sin \theta+3}\right|+ C$
$A =\frac{1}{5}$ and $B (\theta)=\frac{2 \sin \theta+1}{\sin \theta+3}$