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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 :

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

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

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

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

Correct Option: , 4

Solution:

Let $\sin \theta=t \Rightarrow \cos \theta d \theta=d t$

$\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^{2} \theta} d \theta=\frac{d t}{5+7 t-2+2 t^{2}}$

$\Rightarrow \frac{1}{2} \int \frac{d t}{\left(t+\frac{7}{4}\right)^{2}-\left(\frac{5}{4}\right)^{2}}=\frac{1}{5} \ln \left|\frac{t+\frac{1}{2}}{t+3}\right|+C$

$=\frac{1}{5} \ln \left|\frac{2 t+1}{t+3}\right|+C=\frac{1}{5} \ln \left|\frac{2 \sin \theta+1}{\sin \theta+3}\right|+C$

$\therefore B(\theta)=\frac{2 \sin \theta+1}{2(\sin \theta+3)}$ and $A=\frac{1}{5}$

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