Question

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

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

$I=\int \frac{\cos \theta }{5+7\sin \theta -2+2{\sin }^2\theta }d\theta$
Put $\sin \theta =t$,
$\cos \theta d\theta =dt$
$I=\int \frac{dt}{2t^2+7t+3}$
$I=\frac{1}{2}\int \frac{dt}{t^2+\frac{7t}{2}+\frac{3}{2}}=\frac{1}{2}\int \frac{dt}{t^2+\frac{7}{2}t+{\left(\frac{7}{4}\right)}^2-\frac{49}{16}+\frac{24}{16}}$
$I=\frac{1}{2}\int \frac{dt}{{(t+\frac{7}{4})}^2-{\left(\frac{5}{4}\right)}^2}$
$I=\frac{1}{2}\times \frac{1}{2\times \frac{5}{4}}\ln \left|\left[\frac{t+\frac{7}{4}-\frac{5}{4}}{t+\frac{7}{4}+\frac{5}{4}}\right]\right|+C$
$I=\frac{1}{5}\ln \left|\left[\frac{t+\frac{1}{2}}{t+3}\right]\right|+C$
Putting value of $t$ back
$I=\frac{1}{5}\ln \left|\left(\frac{\sin \theta +1/2}{\sin \theta +3}\right)\right|+C$
On comparing we get,
$\frac{B\left(\theta \right)}{A}=\frac{\left(\frac{2\sin \theta +1}{\sin \theta +3}\right)}{\frac{1}{5}}$
$\frac{B\left(\theta \right)}{A}=5\left(\frac{2\sin \theta +1}{\sin \theta +3}\right)$