Question

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