## Calculus: Early Transcendentals 8th Edition

Answer is in an implicit form : $\theta sin\theta + cos\theta = -e^{-t^2}/2 + C$
$d\theta /dt = tsec\theta /(\theta e^{t^{2}})$ 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. $\theta sin\theta + cos\theta = -e^{-t^2}/2 + C$