Not sure how to calculate $\int_0^1 \frac 1 {1+y\cos(x)}dx$ $$\int_0^1 \frac 1 {1+y\cos(x)}dx$$
If there was no $y$, I would multiply by $1-\cos(x)$ and finish it quickly. But the $y$ stops me from doing that. I tried the trigonometric substitution but failed. I assume there is a simple way to solve this. I would appreciate if anyone could help me with this. Thanks.
EDIT: When trying the Weirstrass Sub. I got here and wasn't able to find a way to go forward:
$$\int_0^{\pi/4}\frac {2dt}{1+y+t^2(1-y)}$$
 A: Hint:


*

*Substituting $t = \tan \frac{x}{2}$ gives
$$I(y):=\int \frac{2}{1+y\cos x} \; \mathrm d x= \int \frac{2}{1+y \frac{1-t^2}{1+t^2}} \cdot \frac{2 \; \mathrm d t}{1+t^2} = \int \frac{2}{1+y+t^2(1-y)} \; \mathrm d t .$$

*Since 
$$\int \frac{1}{1+t^2} \; \mathrm d t = \arctan t$$
the final substitution should be easy.

*The last integral is of the type
$$ \int \frac{1}{a +bt^2} \; \mathrm d t $$
with constants $a,b \in \mathbf R$. Now substitute $\frac{\sqrt{b}}{\sqrt{a}}t= z$ and you will get
$$ \int \frac{1}{a +bt^2} \; \mathrm d t = \frac{1}{\sqrt{a}\sqrt{b}} \; \arctan \left( \frac{\sqrt{b}}{\sqrt{a}}t \right) .$$
A: First one should be careful to consider only $y$ such that $1+y\cos x\neq 0$. 
If that is the case we notice that
$$
\frac{1}{1+y\cos x}=\frac{1-y\cos x}{1-y^2\cos^2x}.
$$
Next, we divide into the parts
$$
\frac{1}{1-y^2\cos^2x}=\frac{1}{\sin^2 x+(1-y^2)\cos^2x}=\frac{1}{\cos^2x}\frac{1}{\tan^2x+1-y^2}
$$
For this part, let $u=\tan x$. For the second part,
$$
\frac{-y\cos x}{1-y^2\cos^2x}=-\frac{y \cos x}{1-y^2+y^2\sin^2x}.
$$
Here, let $u=\sin x$.
I think you are now on safe ground. Ask if you cannot take it from here.
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\begin{align}
\int_{0}^{1}{\dd x \over 1 + y\cos\pars{x}} & =
\int_{0}^{1}{\dd x \over 1 + y\bracks{2\cos^{2}\pars{x/2} - 1}} =
2\int_{0}^{1/2}{\dd x \over 1 - y + 2y\cos^{2}\pars{x}}
\\[5mm] & =
2\int_{0}^{1/2}{\sec^{2}\pars{x} \over \pars{1 - y}\sec^{2}\pars{x} + 2y}\,\dd x =
2\int_{0}^{1/2}
{\sec^{2}\pars{x} \over \pars{1 - y}\tan^{2}\pars{x} + 1 + y}\,\dd x
\\[5mm] & =
2\,{1 \over 1 + y}\,\root{1 + y \over 1 - y}\int_{0}^{1/2}
{\root{\pars{1 - y}/\pars{1 + y}}\sec^{2}\pars{x} \over
\bracks{\root{\pars{1 - y}/\pars{1 + y}}\tan\pars{x}}^{2} + 1}\,\dd x
\\[5mm] & =
{2 \over \root{1 - y^{2}}}\,
\arctan\pars{\root{1 - y \over 1 + y}\tan\pars{1 \over 2}}
\end{align}
