I have to show that the following the integral in terms of $x$ can be rewritten as follows:

$$\int_{-1}^1\frac{\sqrt{1-x^2}}{1+x^2}\ \mathrm dx = \int_{u_2=}^{u_1=}\frac1{1+cos^2u}\ \mathrm du-\pi$$

I have used the substitution $x=\cos u$

My working is as follows, I'm just struggling to manage to get the $-\pi$ term:

$$\begin{align} &\int_{\pi}^{0}\frac{\sqrt{1-\cos^2u}}{1+\cos^2u}\cdot-\sin u\ \mathrm du\\[5pt] &\int_{\pi}^{=0}\frac{-\sin^2 u}{1+\cos^2u}\ \mathrm du\\[5pt] &\int_{0}^{\pi}\frac{\sin^2 u}{1+\cos^2u}\ \mathrm du \end{align}$$

  • $\begingroup$ You don't need to write $u_1=$ in the limits. $\endgroup$ – Ben Longo Jan 1 '16 at 22:16
  • $\begingroup$ $1/(1+\cos^2 u)=2/(3+\cos 2u)$. $\endgroup$ – DanielWainfleet Jan 1 '16 at 22:48

From your last step you can write $$I=\int_0^{\pi}\frac{1-\cos^2u}{1+\cos^2u}du=\int_0^{\pi}\frac{-1-\cos^2u}{1+\cos^2u}+\frac{2}{1+\cos^2u}du$$

However $\frac{2}{1+\cos^2u}$ is an even function. Therefore, $$I=-\pi+\frac 12\int_{-\pi}^{\pi}\frac{2}{1+\cos^2u}du$$

And hence the result.


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