Answer
$-\dfrac{1}{\sqrt 2} \csc (\sqrt 2 \theta )+c$
Work Step by Step
Calculate the anti-derivative:
Since, we know $\int \csc x \cot x =-\csc x+C$
Substitute $k=\sqrt 2 \theta \implies dk=\sqrt 2 d \theta$
$=\dfrac{1}{\sqrt 2} \int \csc k \cot k dk$
or, $=-\dfrac{1}{\sqrt 2} \csc k+c$
Back substitution: $k=\sqrt 2 \theta $
$=-\dfrac{1}{\sqrt 2} \csc (\sqrt 2 \theta )+c$
