University Calculus: Early Transcendentals (3rd Edition)

$y(t)=-\cos (e^t-2) +1$
Given: $\dfrac{dy}{dt}=e^t \sin (e^t-2)$ Here, $y=\int e^t \sin (e^t-2) dt$ Plug $a=e^t-2 \implies da=e^t dt$ Now, we have $y=\int \sin a da$ or, $y=-\cos a da+c$ Thus, $y=-\cos (e^t-2 )+c$ Apply initial conditions $y[\ln (2)]=2$ Thus, we have $y(t)=-\cos (e^t-2) +1$