Answer
$$\int \frac{1}{1+cos^{2}\theta}d\theta=\frac{1}{\sqrt{2}}arctan(\frac{tan\theta}{\sqrt{2}})+C$$
Work Step by Step
$$\int \frac{1}{1+cos^{2}\theta}d\theta=\int \frac{1}{1+cos^{2}\theta}\frac{sec^{2}\theta}{sec^{2}\theta}d\theta$$
$$=\int\frac{sec^{2}\theta}{sec^{2}\theta+1}d\theta=\int \frac{sec^{2}\theta}{tan^{2}\theta+2}d\theta$$
$$=\int \frac{1}{tan^{2}\theta+2}d(tan\theta)$$
$$=\frac{1}{\sqrt{2}}arctan(\frac{tan\theta}{\sqrt{2}})+C$$
