Evaluare $I(y)=\int_0^1\frac{1}{y+\cos(x)}dx$. Hence determine $J(y)=\int_0^1\frac{1}{(y+\cos(x))^2}dx$ 
Consider the following integral:
$$I(y)=\int_0^1\frac{1}{y+\cos(x)}dx $$
with Weierstrasse substitution I showed that
$$ I(y)=\frac{2}{\sqrt{y^2-1}}\arctan\left(\sqrt{\frac{{y-1}}{y+1}}\right). $$
It is this next part that I am not sure about: Hence detertemine
$$J(y)=\int_0^1\frac{1}{(y+\cos(x))^2}.$$

My attempt:
Since $I(y)\pm J(y)$ yields no simplication, the most obvious approach would be to rewrite $I(y)$ as
$$I(y)=\int_0^1\frac{(y+\cos(x))}{(y+\cos(x))^2}=yJ(y)+\underbrace{\int_0^1\frac{cos(x)}{(y+\cos(x))^2}dx}_{K(y)}$$
Applyling Weierstrasse substitituion for $K(y)$:
$$K(y)=2\int_0^1\frac{(1-x^2)}{((y-1)x^2+(y+1))^2}dx=-2\int_0^1\frac{\frac{1}{y-1}((y-1)x^2+(y+1))-\frac{y+1}{y-1}-1}{((y-1)x^2+(y+1))^2}$$
$$=-2\left(\underbrace{\frac{1}{y-1}\int_0^1\frac{1}{(y-1)x^2+(y+1)}dx}_{\text{elementary integral}\to \arctan()}-\underbrace{\frac{2}{y-1}\int_0^1\frac{1}{((y-1)x^2+(y+1))^2}dx}_{\text{evaluated with } 
 \tan(u)=\sqrt{\frac{y-1}{y+1}}x}\right)$$
so what follows is quite elmenatary.

However it seems to me that $K(y)$ is by no means simpler than $J(y)$, (i.e. the same approach can be used for $J(y)$ but without having to use the result for $I(y)$) so this is clearly not the point of the question?
 A: As @JackD'Aurizio pointed out, we trying to find:
$$\frac{\partial}{\partial\text{n}}\left(\frac{1}{\text{n}+\cos\left(x\right)}\right)\tag1$$
Using the chain rule, we get:
$$\frac{\partial}{\partial\text{n}}\left(\frac{1}{\text{n}+\cos\left(x\right)}\right)=-\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\cdot\frac{\partial}{\partial\text{n}}\left(\text{n}+\cos\left(x\right)\right)\tag2$$
Differentiate the sum term by term:
$$\frac{\partial}{\partial\text{n}}\left(\frac{1}{\text{n}+\cos\left(x\right)}\right)=-\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\cdot\left(\frac{\partial}{\partial\text{n}}\left(\text{n}\right)+\cos\left(x\right)\cdot\frac{\partial}{\partial\text{n}}\left(1\right)\right)=$$
$$-\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\cdot\left(1+\cos\left(x\right)\cdot0\right)=-\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\tag3$$
So, we have:
$$\int_0^1\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\space\text{d}x=-\int_0^1\frac{\partial}{\partial\text{n}}\left(\frac{1}{\text{n}+\cos\left(x\right)}\right)\space\text{d}x=$$
$$-\frac{\partial}{\partial\text{n}}\left(\int_0^1\frac{1}{\text{n}+\cos\left(x\right)}\space\text{d}x\right)\tag4$$
A: I suppose that the trick is
$$\frac{1}{(y+\cos (x))^2}=\frac{y}{\left(y^2-1\right) (y+\cos (x))}-\frac{y \cos (x)+1}{\left(y^2-1\right) (y+\cos (x))^2}$$
