Calculate the integral of $ frac{1}{1 + x^2} $ over the unit circle

To calculate the integral of $ \frac{1}{1 + x^2} $ over the unit circle, we first convert to polar coordinates:

$ x = r \cos(\theta), y = r \sin(\theta) $

In polar coordinates, the unit circle is defined as:

$ r = 1 $

Substituting in the integral:

$ \int_0^{2\pi} \frac{r}{1 + r^2 \cos^2(\theta)} d\theta $

Since $r = 1$:

$ \int_0^{2\pi} \frac{1}{1 + \cos^2(\theta)} d\theta $

Applying the Weierstrass substitution:

Let $ \tan(\theta/2) = t $, then $ d\theta = \frac{2}{1+t^2} dt $

The integral becomes:

$ \int_{-\infty}^{\infty} \frac{2}{1 + \cos^2(2 \arctan(t))} \frac{1}{1+t^2} dt $

After simplification, the integral reduces to:

$ \pi \int_{-\infty}^{\infty} \frac{2}{2 + t^2} \frac{1}{1+t^2} dt $

The final answer is:

$ \pi \ln{2} $

Convert to polar coordinates:

$ x = r cos( heta), y = r sin( heta) $

Unit circle: $ r = 1 $

Substitute:

$ int_0^{2pi} frac{r}{1 + r^2 cos^2( heta)} d heta $

Since $r = 1$:

$ int_0^{2pi} frac{1}{1 + cos^2( heta)} d heta $

Weierstrass substitution:

$ an( heta/2) = t $, $ d heta = frac{2}{1+t^2} dt $

Integral becomes:

$ int_{-infty}^{infty} frac{2}{1 + cos^2(2 arctan(t))} frac{1}{1+t^2} dt $

Simplified:

$ pi int_{-infty}^{infty} frac{2}{2 + t^2} frac{1}{1+t^2} dt $

Final answer:

$ pi ln{2} $

Convert to polar coordinates:

$ x = r cos( heta), y = r sin( heta) $

Unit circle: $ r = 1 $

Substitute:

$ int_0^{2pi} frac{1}{1 + cos^2( heta)} d heta $

Weierstrass substitution:

$ an( heta/2) = t $, $ d heta = frac{2}{1+t^2} dt $

Simplified integral:

$ pi int_{-infty}^{infty} frac{2}{2 + t^2} frac{1}{1+t^2} dt $

Final answer:

$ pi ln{2}

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