# 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.

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I can't post images, but I've left it in latex. You can view it on codecogs.com/latex/eqneditor.php –  Joe Francis Nov 22 '12 at 22:38
Went there, saw nothing. –  Gerry Myerson Nov 22 '12 at 22:41
Try $x=\tan\theta$. –  Gerry Myerson Nov 22 '12 at 22:43
@JoeFrancis I typeset your equation to appear properly. Are the equations rights? If so then both these integrals are different and give different values. –  user17762 Nov 22 '12 at 22:45
@JoeFrancis : Notice that $\dfrac{\sqrt{A}}{A}$ $=\dfrac{\sqrt{A}}{\sqrt{A}\sqrt{A}}$ $=\dfrac{1}{\sqrt{A}}$. So $\dfrac{\sqrt{1+x^2}}{1+x^2}$ $=\dfrac{1}{\sqrt{1+x^2}}$. –  Michael Hardy Nov 23 '12 at 1:03

## 1 Answer

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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