Question

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

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

Answer 1

The correct option is A

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

Explanation for the correct option.

Find the value of $\frac{B\left(\theta \right)}{A}$:

Given,

$\int \frac{\cos \theta }{5+7\sin \theta -2{\cos }^2\theta }d\theta =A{\log }_e\left|B\left(\theta \right)\right|+C$.

Let, $\sin \theta =t$

$\Rightarrow \cos \theta d\theta =dt$

So, the integral should be,

$\begin{array}{rcl}\int \frac{\cos \theta }{3+7\sin \theta +2\left(1-{\cos }^2\theta \right)}d\theta & =& A{\log }_e\left|B\left(\theta \right)\right|+C\\ & \Rightarrow & \int \frac{\cos \theta }{3+7\sin \theta +2{\sin }^2\theta }d\theta =A{\log }_e\left|B\left(\theta \right)\right|+C\\ & \Rightarrow & \int \frac{dt}{3+7t+2t^2}=A{\log }_e\left|B\left(\theta \right)\right|+C\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{=}\mathbf{t}\mathbf{}\mathbf{\&}\mathbf{\&}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{d}\mathbf{\theta }\mathbf{=}\mathbf{d}\mathbf{t}\mathbf{]}\\ & \Rightarrow & \frac{1}{5}\int \frac{5dt}{(2t+1)(t+3)}=A{\log }_e\left|B\left(\theta \right)\right|+C\\ & \Rightarrow & \frac{1}{5}\int \left(\frac{2}{2t+1}-\frac{1}{t+3}\right)dt=A{\log }_e\left|B\left(\theta \right)\right|+C\\ & \Rightarrow & \frac{1}{5}{\log }_e\left|\frac{2t+1}{t+3}\right|+C=A{\log }_e\left|B\left(\theta \right)\right|+C\\ & \Rightarrow & \frac{1}{5}{\log }_e\left|\frac{2\sin \theta +1}{\sin \theta +3}\right|+C=A{\log }_e\left|B\left(\theta \right)\right|+C\end{array}$

On comparing the coefficient of both side,

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

Therefore,

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

Hence, the correct option is A.