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

Answer 1

Answer: (B)

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