<s>[INST] Tricky integration by substitution $\int_{-1}^{1} \frac{ \sqrt{1-x^2}}{1+x^{2}} dx$ I have to get this integral
$$\int_{-1}^{1} \frac{ \sqrt{1-x^2}}{1+x^{2}} dx$$
into
$$\int_{-\pi }^{\pi } \frac{1}{1+\cos^2\theta } \,d\theta - \pi$$  
any tips would be recommended.
 [/INST] I will use the substitution $x = \cos θ$. And my result is very close to the required.
$dx = - \sinθ d \theta $
When $x = 1, \ θ=\dots  = 0$ and when $x = –1, \ θ= \dots = – π.$
We have : 
$$ \begin{align}
& \int_{- \pi}^{0} \cfrac{ \sqrt{1- \cos^2 \theta} }{1+ \cos^{2} \theta} (- \sin \theta d \theta) \\ 
& = – \int_{-\pi}^{0} \cfrac{\sin^2 \theta}{1+ \cos^2 \theta} d \theta \\
& = – \int_{- \pi}^{0} \cfrac{1 – \cos^2 \theta}{1+ \cos^2 \theta} d \theta \\ 
& = – \int_{- \pi}^{ 0} \left( –1 + \cfrac {2}{1 + \cos^2 \theta} \right) d \theta \\ 
& = \int_{- \pi}^{0} d \theta  – 2\int_{- \pi}^{\theta} \left( \cfrac {1}{1 + cos^{2} \theta} \right) d \theta \\
& = \pi –2\int_{- \pi}^{0} \left( \color{blue}{\cfrac {1}{1 + cos^{2} \theta}} \right) d \theta  \\
& = \pi –\int_{-\pi}^{\pi}\left( \cfrac {1}{1 + cos^{2} \theta} \right) d \theta \end{align} $$ 
Area under [-π, π] = 2 * that under [-π. 0] (in blue)
Which differs from the requested form by a sign only.
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