Integrate $\frac{1}{1+\cos^2x}$. Probably need using some trigonometric identity I don't know

Question

Integrate $\frac{1}{1+\cos^2x}$

I probably need using some trigonometric identity I don't know. I tried all methods I'm familiar with. Any assistance will be great.

Thank you!

Answer 1

Let $u=\tan x$ then $du=(1+u^2)dx$ so

$$\int \frac{dx}{1+\cos^2x}=\int\frac{du}{(1+u^2)(1+(1+u^2)^{-1})}=\int\frac{du}{2+u^2}$$ Can you take it from here?

Answer 2

Setting $$ t=\tan(x) $$ we get $$ \frac{1}{1+\cos^2x}=\frac{\cos^2x+\sin^2x}{2\cos^2x+\sin^2x}=\frac{1+t^2}{2+t^2}. $$ Since $$ t=\tan x\iff x=\tan^{-1}t, $$ and $$ dx=\frac{dt}{1+t^2}, $$ it follows that \begin{eqnarray} \int\frac{dx}{1+\cos^2x}&=&\int\frac{1+t^2}{2+t^2}\cdot\frac{dt}{1+t^2}=\int\frac{dt}{2+t^2}=\frac{1}{\sqrt2}\tan^{-1}\left(\frac{t}{\sqrt2}\right)+A\\ &=&\frac{1}{\sqrt2}\tan^{-1}\left(\frac{\tan x}{\sqrt2}\right)+A, \end{eqnarray} with $A$ an arbitrary real constant.

Answer 3

I know you already have an answer but you could divide top and bottom by $\cos^2(x)$ So you have $\int \frac{\sec^2(x)}{2+\tan^2(x)} dx$ Let $u=\tan(x)$ But this looks similar to @math 's idea. Oh by the way I used $ \sec^2(x)=\tan^2(x)+1$