Answer
$\displaystyle \theta \sin(\theta) +\cos(\theta) =-\frac{1}{2e^{t^2}}+C$
Work Step by Step
$\displaystyle \frac{d\theta}{dt}=\frac{t\sec(\theta)}{\theta e^{t^2}}=\frac{t}{\theta\cos(\theta)e^{t^2}}$
Separation of variables:
$\displaystyle \int\theta\cos(\theta)d\theta = \int\frac{t}{e^{t^2}}dt$
Integrate with respect to each variable:
$\qquad\qquad\qquad$Left side:
Integration by parts: $\displaystyle \int f\cdot g'dx = fg - \int f'\cdot gdx$
Let $ f = \theta$ and $g' = \cos(\theta)$
$\therefore f = 1\qquad g = \sin(\theta)$
$\displaystyle \int\theta\cos(\theta)d\theta = \theta \sin(\theta) - \int(1)\sin(\theta)d\theta$
$\qquad\qquad\qquad = \theta \sin(\theta) - [-\cos(\theta)]$
$\qquad\qquad\qquad = \theta \sin(\theta) +\cos(\theta) $
$\qquad\qquad\qquad$ Right side:
$\displaystyle \int\frac{t}{e^{t^2}}dt = \int te^{t^{-2}}\qquad$let $u = t^{-2}\quad\rightarrow\quad du = -2t dt \quad\rightarrow\quad -\frac{1}{2}du = t dt$
$\displaystyle -\frac{1}{2}\int e^udu=-\frac{1}{2}e^u+C$
$\displaystyle -\frac{1}{2}e^{t^{-2}}+C=-\frac{1}{2e^{t^2}}+C$
Set the two sides equal to each other again:
$\displaystyle \theta \sin(\theta) +\cos(\theta) =-\frac{1}{2e^{t^2}}+C$
