How to evaluate $\int \frac {dx}{\cos x+C}$?

Question

I want to evaluate -

$$\int \frac {dx}{\cos x+C}$$

Where $C$ is an arbitrary constant.I tried substitution and parts but could not do it.

Note that for $C=1$ one can simply do this by using compound angle formulas.But what about other values of $C $?

Thanks for any help!!

Answer 1

Hint: use the substitution $t=\tan{\frac{1}{2}x}$. Then $dx = 2dt/(1+t^2)$, $$ \cos{x} = \frac{1-t^2}{1+t^2}, $$ and the integral reduces to that of a fairly simple rational function.

Answer 2

Hint: Make the substitution $ \theta = \frac{x}{2}$.

\begin{align} \frac{1}{C + \cos 2 \theta} &= \frac{1}{C -1 + 2\cos ^2 \theta} \\ &= \frac{\sec^2 \theta}{(C-1)\sec^2 \theta + 2} \\ &= \frac{\sec^2 \theta}{C + 1 + (C-1)\tan^2 \theta} \end{align} Hence

\begin{align} \int \frac{1}{C + \cos x} \text{d}x &= \int \frac{2}{C + \cos 2 \theta} d \theta \\ &= \int \frac{ 2\sec^2 \theta}{ C + 1 + (C-1)\tan^2 \theta} d \theta \end{align}

Now make the second substitution $t = \tan \theta$ and continue as above.