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My approach was to use a double-arc formula for cosine and sine, but I did not succeed. $\sin^2 x = \frac{1 - \cos(2x)}{2}$ and $\cos^2x = \frac{1 + \cos(2x)}{2}$

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  • $\begingroup$ I would try using $1=\cos^2 x+\sin^2 x$ then make an obvious substitution. $\endgroup$ – Shaun Jul 15 '18 at 22:23
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We reduce the degree of the numerator with

$$\int\frac{\cos^2x}{1+\sin^2x}dx=\int\frac{2-(1+\sin^2x)}{1+\sin^2x}dx=2\int\frac{dx}{1+\sin^2x}-x$$

and we introduct a tangent to rationalize,

$$\int\frac{dx}{1+\sin^2x}=\int\frac{dx}{\cos^2x+2\sin^2x}=\int\frac{dx}{\cos^2x(1+2\tan^2x)}=\int\frac{dt}{1+2t^2} \\=\frac1{\sqrt2}\int\frac{du}{1+u^2}=\frac1{\sqrt2}\arctan u=\frac1{\sqrt2}\arctan\sqrt2t=\frac1{\sqrt2}\arctan(\sqrt2\tan x).$$

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BE CAREFUL:

$\frac{\cos^2(x)}{1+\sin^2(x)}$ is a continuous function over $\mathbb{R}$, therefore you would generally want an antiderivative which also has that property. It is here that WolframAlpha is evidently inferior to GeoGebra (with integration) as latter gives the wonderful solution:

$$ \int \frac{\cos^2(x)}{1+\sin^2(x)} dx = \sqrt{2} \arctan \left( {\frac{(2-\sqrt2)\sin2x}{(\sqrt2 - 2)\cos2x +\sqrt2 +2}} \right) +(\sqrt2 -1)x +C$$

By graphical inspection (or otherwise), after implementing the Weierstrass substitution, one can deduce that the continuous integral is

$$ \sqrt2 \arctan(\sqrt2 \tan(x)) -x + k \left\lfloor \frac{x+\pi/2}{\pi} \right\rfloor + C $$ Where, $f(x) := \sqrt2 \arctan(\sqrt2 \tan(x)) -x $, $$k = \lim_{x \to \pi/2 ^-} f(x) - \lim_{x \to \pi/2 ^+} f(x) $$

Then, to retrieve GeoGebra's beautiful closed form solution, use: $$ \lfloor x \rfloor = \frac{1}{\pi}(\arctan(\cot(\pi x))+\pi x-\pi/2)$$

Read Szeto's answer here: Continuous antiderivative of $\frac{1}{1+\cos^2 x}$ without the floor function. for the floor function insight and concerns about discontinuity.

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Bioche's rules suggest the substitution $\; t=\tan x$, so $\mathrm d t=(1+t^2)\,\mathrm dx\iff \mathrm dx=\dfrac{\mathrm d t}{1+t^2}$.

Indeed, $\;\cos^2x=\dfrac 1{1+t^2}$, $\;\sin^2x=t^2\cos^2x=\dfrac {t^2}{1+t^2}$, so that $$\int \frac{\cos^2 x}{1 + \sin^2 x}\,\mathrm dx=\int \frac{\mathrm d t}{(1 + t^2)(1 + 2t^2)}$$ There remains to decompose this fraction into partial fractions: $$ \frac{1}{(1 + t^2)(1 + 2t^2)}= \frac{A}{(1 + t^2)}+\frac{B}{(1 + 2t^2)}.$$ Can you proceed from there?

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  • $\begingroup$ Do you have a source for Bioche's rule? I am interested in it, but I did not find anything usefull yet. Just fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche, but I do not speak french. $\endgroup$ – Cornman Jul 16 '18 at 7:03
  • $\begingroup$ I've found this online Google excerpt of a book by Sean M. Stewart, which explains what I learnt when I was a student. $\endgroup$ – Bernard Jul 16 '18 at 7:30
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HINT Proceed with the following substitution $\tan\frac{x}{2}=t$, then $x=2\arctan t$, $dx=2dt/(1+t^2)$.

Remember that $\sin x=\frac{2t}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$.

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Try this $$\int \dfrac{\cos^2x}{1+\sin^2x}dx=\int\dfrac{1-\sin^2x}{1+\sin^2x}dx=\int \biggl(\dfrac{2}{1+\sin^2x}-1\biggr)dx$$ Can you take it from here?

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Note that we can write

$$\begin{align} \frac{\cos^2(x)}{1+\sin^2(x)}&=\frac{1+\cos(2x)}{3-\cos(2x)}\\\\ &=\frac{4-(3-\cos(2x))}{3-\cos(2x)}\\\\ &=-1+\frac{4}{3-\cos(2x)} \end{align}$$

Enforce the substitution $y=\tan(x)$, so that $\cos(2x)=\frac{1-y^2}{1+y^2}$ and $dx=\frac1{1+y^2}\,dy$. Proceeding reveals

$$\begin{align} \int\frac{4}{3-\cos(2x)}\,dx&=\int \frac{2}{1+2y^2}\,dy\\\\ &=\sqrt2\arctan(\sqrt2\,y)+C\\\\ &=\sqrt{2}\arctan(\sqrt2 \tan(x))+C \end{align}$$

Putting it all together yield

$$\int \frac{\cos^2(x)}{1+\sin^2(x)}\,dx=-x+\sqrt{2}\arctan(\sqrt2 \tan(x))+C$$

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Jul 21 '18 at 18:15
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Use the substitution: $$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$dx=\frac{2t}{1+t^2}dt$$the so called Weierstrass Substitution.

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