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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.

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    @limac246, great catch! Tha$n$ks.2012-11-23

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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.