## Calculus (3rd Edition)

$$y= \frac{1}{\sec x }\left( x +C\right).$$
This is a linear equation which has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{ \int \tan x dx}=e^{-\ln(\cos x)}=\sec x$$ where we used the fact that $\int \tan xdx=-\ln(\cos x)$. Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=\frac{1}{\sec x }\left( \int \sec x \cos xdx +C\right)\\ &= \frac{1}{\sec x }\left( \int dx +C\right)\\ &= \frac{1}{\sec x }\left( x +C\right)\\ \end{align} Thus, the general solution is $$y= \frac{1}{\sec x }\left( x +C\right).$$