$\displaystyle \int{\frac{1}{1+cos^2x}d(x)}$ In this case i divide numerator and denominator by $\displaystyle sec^2x$ and i get $\displaystyle \int{\frac{sec^2x}{1+sec^2x}}{$ like this.Then What can i do???????
Well, you'd actually be MULTIPLYING the top and bottom by $\displaystyle \displaystyle \begin{align*} \sec^2{(x)} \end{align*}$ to give you this. Anyway...
$\displaystyle \displaystyle \begin{align*} \int{ \frac{\sec^2{(x)}}{1 + \sec^2{(x)}}\,dx} &= \int{ \frac{\sec^2{(x)}}{ 1 + \left[ 1 + \tan^2{(x)} \right] } \, dx} \\ &= \int{ \frac{\sec^2{(x)}}{ 2 + \tan^2{(x)}}\,dx} \end{align*}$
Now make the substitution $\displaystyle \displaystyle \begin{align*} \tan{(x)} = \sqrt{2}\tan{(\theta)} \implies \sec^2{(x)}\,dx &= \sqrt{2}\sec^2{(\theta)}\,d\theta \end{align*}$ and the integral becomes
$\displaystyle \displaystyle \begin{align*} \int{ \frac{\sec^2{(x)}}{2 + \tan^2{(x)}}\,dx} &= \int{ \frac{\sqrt{2}\sec^2{(\theta)}}{2 + \left[ \sqrt{2} \tan{(\theta)} \right] ^2 } \, d\theta} \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{ 2 + 2\tan^2{(\theta)}}\,d\theta} \\ &= \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\theta)}}{1 + \tan^2{(\theta)}}\,d\theta} \\ &= \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\theta)}}{\sec^2{(\theta)}}\,d\theta } \\ &= \frac{\sqrt{2}}{2} \int{ 1\,d\theta} \\ &= \frac{\sqrt{2}}{2} \theta + C \\ &= \frac{\sqrt{2}}{2} \arctan{ \left[ \frac{\sqrt{2}\tan{(\theta)}}{2} \right] } + C \end{align*}$