Answer
Answer is in an implicit form :
$\theta sin\theta + cos\theta = -e^{-t^2}/2 + C$
Work Step by Step
$d\theta /dt = tsec\theta /(\theta e^{t^{2}})$
Now let's separate $\theta$ on one side and t on the other side:
$(\theta / sec\theta ) d\theta = t/ e^{t^{2}} dt $
It is to be noted that $ sec \theta = 1/cos\theta$
The equation can then be written as:
$(\theta cos\theta ) d\theta = t/ e^{t^{2}} dt $
Now let's integrate both sides:
$\int(\theta cos\theta ) d\theta =\int t/ e^{t^{2}} dt $
Left-hand side:
Use integration by parts to solve using $u = \theta$ and $v' = cos \theta$
We will have :
$\int(\theta cos\theta ) d\theta $
$= \theta sin\theta - \int sin\theta d\theta = \theta sin\theta + cos\theta + C_1 $
Right-hand side
Use substitution to solve the equation using $z = t^2$ so $dz/dt = 2t$
$\int t/ e^{t^{2}} dt $
$=\int e^{-z} dz/2$
$ = -e^{-z}/2 + C_2$
$ = -e^{-t^2}/2 + C_2$
Now, let's put both sides together and we can combine $C_1$ and $C_2$ to make a C.
$\theta sin\theta + cos\theta = -e^{-t^2}/2 + C$
