y' + ytanx = secx Solve the first-order differential equation

Given y'+y tanx = secx
when the first order linear ordinary Differentian equation has the form of
y'+p(x)y=q(x)
then the general solution is ,
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)
so,
y'+y tanx = secx--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = tanx and q(x)=sec x
so on solving with the above general solution we get:
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)
=((int e^(int tanx dx) *(secx)) dx +c)/e^(int tanx dx)
first we shall solve
e^(int tanx dx)=e^(ln(secx)) = sec x
as we knowint tanx dx = ln(secx)  
So, proceeding further, we get
y(x) =((int secx *secx) dx +c)/secx
=(int sec^2 x  dx +c)/secx
=(tanx+c)/secx
=tanx/secx + c/secx
y(x)=sinx+ c*cosx