Solve the differential equation.
-$$\frac { d y } { d x } = 3 \cos x \sec y$$
A) $y = 3 \sin x + C$
B) $y = \sin ^ { - 1 } ( 3 \cos x + C )$
C) $y = \sin ( 3 \sin x + C )$
D) $y = \sin ^ { - 1 } ( 3 \sin x + C )$

Question

Quizplus · Accepted Answer

We can separate the variables and write:
$$\sec y dy = 3 \cos x dx$$
Taking the antiderivative of both sides, we get:
$$\ln|\sec y + \tan y| = 3\sin x + C_1$$
where $C_1$ is the constant of integration.

Applying $\sin^{-1}$ to both sides, we have:
$$y = \sin^{-1}(\sec y + \tan y) = \sin^{-1}(3\cos x + C)$$
where $C = \sin^{-1}(C_1)$. Therefore, the answer is $\boxed{\textbf{(D)}}$.

Solve the Differential Equation dydx=3cos⁡xsec⁡y\frac { d y } { d x } = 3 \cos x \sec ydxdy​=3cosxsecy

Solve the Differential Equation $\frac { d y } { d x } = 3 \cos x \sec y$