I have been trying to solve the ODE
$$\frac {du}{dx} = 1 + \cos u. $$
To solve it, I divided through by $1 + \cos u$ to give
$$\frac{1} {1 + \cos u } \frac{du}{dx} = 1,$$ and then tried to integrate both sides.
However, I have gotten quite stuck when trying to evaluate the integral of $1/ (1 + \cos u )$.
I tried multiplying the integrand by $\dfrac{\cos u -1} {\cos u -1}$, but that led to the integral becoming
$$ \int \frac{\tan u} {\sin^2u} \, du – \int du$$
and I didn't know how to evaluate the first integral,
$$\int \frac{\tan u} {\sin^2u} \,du.$$
So, I would really appreciate it if someone could tell me how to go about solving this ODE. Thanks!
Best Answer
Your algebra isn't correct so far. Multiplying the integrand by the factor you've listed actually gives
$$\int\frac{\cos{u} - 1}{\cos^2{u} -1} du = \int\frac{\cos{u} - 1}{-\sin^2{u}} du$$
which can then be separated as
$$\int\frac{1}{\sin^2{u}} du - \int \frac{\cos{u}}{\sin^2{u}} du$$
The first integral is the same as integrating $\csc^2{u}$, which gives $-\cot{u}$; the second integral can be done by substituting for $\sin u$.