Question

The integral $I=\displaystyle \int sin \left(2 \theta \right)\left[\frac{1 + \left(cos\right)^{2} ⁡ \theta }{2 \left(sin\right)^{2} ⁡ \theta }\right]d\theta $ simplifies to (where, $c$ is the integration constant)

• A. $\ln |\sin \theta|+\cos \theta+c$
• B. $2 \ln |\sin \theta|-\frac{\sin ^{2} \theta}{2}+c$
• C. $\ln |\sin \theta|-\sin ^{2} \theta+c$
• D. $\ln |\cos \theta|+\cos ^{2} \theta+c$

Answer 1

Answer: (B)

$I=\int \frac{\cos \theta\left(1+(\cos )^{2} \theta\right)}{\sin \theta} d \theta$
Let $\sin \theta=t$
$\Rightarrow \cos \theta d \theta=d t$
$\Rightarrow I=\int \frac{2-t^{2}}{t} d t$
$=2 \ln |t|-\frac{t^{2}}{2}+c$
$=2 \ln |\sin \theta|-\frac{\sin ^{2} \theta}{2}+c$