Evaluating $\displaystyle\int_0^1\frac{\sqrt{1-y^2}}{1+y^2}dy$ without trig substitution A problem in our calculus text requires the evaluation of $\displaystyle\int_0^1\frac{\sqrt{1-y^2}}{1+y^2}dy$, 
and
I have evaluated it by substituting $y=\sin\theta$ (or $y=\tanh u$) and then using another substitution and partial fractions; but I would like to find out if there is a simpler way to find this integral that does not
involve trig substitution (or hyperbolic substitution).
(I know some MSE people dislike this type of question, so I apologize in advance.)
 A: Hint:
$$t=y^2$$
$$dy=\frac{1}{2\sqrt t}dt$$
then
$$I=\frac{1}{2}\int_0^1\frac{\sqrt{\frac{1-t}{t}}}{1+t}dt$$
$$\frac{1-t}{t}=u^2$$
so$$t=\frac{1}{1+u^2} $$
$$I=\int_0^{\infty}\frac{u^2}{(u^2+1)(u^2+2)}du=\int_0^{\infty}(\frac{2}{u^2+2}-\frac{1}{u^2+1})du$$
then you can solve it without trigonometric substitution by knowing that 
$$\int_0^{\infty}\frac{1}{u^2+1}du=\frac{\pi}{2}$$
A: I will try to give an approach to this integral using complex analysis:
First of all we would like to have a form of the integral which is most easily tractable by contour integration, which (at least for me) means that 


*

*it is as obivious as possible to guess a contour which can be used

*the pole/cut structure is as clear as possible
One way of doing so, is exploiting the parity of the integrand and transform $y\rightarrow 1/x$. We get:
$$
I=\frac{1}{2}P\int_{-1}^{1}\frac{1}{x}\frac{\sqrt{x^2-1}}{1+x^2}dx
$$
Here $P$ denotes Cauchy's principal value.
We now may consider the complex function 
$$
f(z)=\frac{1}{z}\frac{\sqrt{z^2-1}}{1+z^2}
$$
Choosing the standard branch of logarithm, we have a cut on the interval $[-1,1]$, furthermore we have singularities at $\{\pm i,0\}$ where the first two are harmless but the one at zero is on the cut and will need to be handeled with care.
Now may choose a contour which encloses the branch cut, and avoids the singularity at 0. we get:
$$
\oint f(z)dz = \underbrace{\int_{-1}^{-\epsilon}f(x_+)+\int_{\epsilon}^{1}f(x_+)}_{2I}+\int_{\text{arg}(z)\in(\pi,0], |z|=\epsilon}f(z)dz\\-\underbrace{\int_{-1}^{-\epsilon}f(x_-)-\int_{\epsilon}^{1}f(x_-)}_{-2I}-\int_{\text{arg}(z)\in[0,-\pi), |z|=\epsilon}f(z)dz=\\
4I+\underbrace{2\int_{\text{arg}(z)\in(\pi,0], |z|=\epsilon}f(z)dz}_{2 \times \pi i\ \times \text{res}[f(z),z=0] }=\\
4I +2\pi \quad (1)
$$
Where $\text{res}[f(z),z=0]=i$ because we calculated the residue above the branch cut. Furthermore $f(x_{\pm})$ denotes about which side of the cut we are talking: $\pm$ above/below. Also the limit $\epsilon \rightarrow 0$ is implicit.
Now comes the trick: 
By looking at the exterior of the contour we can also write  (please note that we now enclose the singularities in opposite direction compared to above)
$$
\oint f(z)dz=-2\pi i \times(\text{res}[f(z),z=i]+
\text{res}[f(z),z=-i])=2\sqrt{2}\pi \quad (2)
$$
Equating $(1)=(2)$
$$
4I+2\pi=2\sqrt{2}\pi\\
$$
or
$$
I=\frac{\pi}{2}\left(\sqrt{2}-1\right)
$$
which is the same result as the one obtained by trig. substitution.
A: Let us take the "gruesome" path.
$$\begin{eqnarray*}\int_{0}^{1}\frac{\sqrt{1-x^2}}{1+x^2}\,dx&=&\sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{2n}\sqrt{1-x^2}\,dx = \frac{\sqrt{\pi}}{4}\sum_{n\geq 0}(-1)^n\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+2)}\\&=&\frac{\pi}{4}\sum_{n\geq 0}\frac{(-1)^n}{4^n(n+1)}\binom{2n}{n}=\left.\frac{\pi}{2}\cdot\frac{1-\sqrt{1-x}}{x}\right|_{x=-1}\\&=&\color{red}{\frac{\pi}{2}(\sqrt{2}-1)}.\end{eqnarray*}$$
Not so painful, after all, if one recalls the generating function of the Catalan numbers.
Steps involved:

*

*expansion of $\frac{1}{1+x^2}$ as a geometric series;

*integration of $x^{\alpha}(1-x)^{\beta}$ over $[0,1]$ through the Euler beta function;

*rewriting in terms of central binomial coefficients;

*evaluation through the generating function for the Catalan numbers.

A: The indefinite integral is
$$
\int\frac{\sqrt{1-y^2}}{1+y^2} dy = -\arcsin(y) - 
\sqrt{2}\arctan\left(\frac{\sqrt{2}\;y}{\sqrt{1-y^2}}\right) + C
$$
so of course it is most easily done with a trig substitution.
Sometimes a definite integral can be done without first doing the indefinite integral.  As another answer suggests,
$$
\int_0^1\frac{\sqrt{1-y^2}}{1+y^2} dy = \frac{1}{2}\int_{-1}^1\frac{\sqrt{1-y^2}}{1+y^2} dy
$$
and that integral is amenable to a solution by residues in the complex plane.
