# Tricky integration by substitution.

I have to get this integral (EDIT: it should definitely be 1-x^2 in numerator) $$\int_{-1}^{1} \frac{ \sqrt{1-x^2}}{1+x^{2}} dx$$ 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