3
$\begingroup$

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.

  • 0
    I can't post images, but I've left it in latex. You can view it on http://www.codecogs.com/latex/eqneditor.php2012-11-22
  • 1
    Went there, saw nothing.2012-11-22
  • 2
    Try $x=\tan\theta$.2012-11-22
  • 2
    @JoeFrancis I typeset your equation to appear properly. Are the equations rights? If so then both these integrals are different and give different values.2012-11-22
  • 0
    There's meant to be a - pi after the dtheta on the second integral2012-11-22
  • 2
    @JoeFrancis : Notice that $\dfrac{\sqrt{A}}{A}$ $=\dfrac{\sqrt{A}}{\sqrt{A}\sqrt{A}}$ $=\dfrac{1}{\sqrt{A}}$. So $\dfrac{\sqrt{1+x^2}}{1+x^2}$ $=\dfrac{1}{\sqrt{1+x^2}}$.2012-11-23
  • 0
    Even after the correction (supposedly?), I'm not seeing that these two integrals are equivalent. Here's what I got for the first: $$ \begin{align} \int_{-\pi}^{\pi}\frac{1}{1+\cos^2\theta}d\theta-\pi&=\int_{-\pi}^{\pi}\frac{1}{1+\cos^2\theta}d\theta-\frac{1}{2}\int_{-\pi}^{\pi}1d\theta\\ &=\int_{-\pi}^{\pi}\frac{1}{1+\cos^2\theta}-\frac{1}{2}d\theta\\ &=\int_{-\pi}^{\pi}\frac{2-(1+\cos^2\theta)}{2(1+\cos^2\theta)}d\theta\\ &=\int_{-\pi}^{\pi}\frac{1+\cos^2\theta}{2(1+\cos^2\theta)}d\theta\\ &=\int_{-\pi}^{\pi}\frac{1}{2}d\theta\\ &=\pi \end{align} $$ This isn't the same as the above integral.2012-11-23
  • 0
    @Limitless, I think you missed a sign in the 3rd to last step as $2- (1+\cos^2 \theta) = 1 - \cos^2 \theta$ not $1 + \cos^2 \theta.$2012-11-23
  • 0
    @limac246, great catch! Thanks.2012-11-23

1 Answers 1