Solve that


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

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

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

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

Correct Option: , 4


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

$\int \frac{\cos \theta d \theta}{3+7 \sin \theta+2 \sin ^{2} \theta} \quad \begin{aligned}&\sin \theta=t \\&\cos \theta d \theta=d t\end{aligned}$

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

$=\frac{1}{5} \int\left(\frac{2}{2 t+1}-\frac{1}{t+3}\right) d t$

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

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

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

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