A problem in our calculus text requires the evaluation of $\int_0^1\frac{\sqrt{1-y^2}}{1+y^2}dy$,
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.)
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
$$\int_0^{\infty}\frac{1}{u^2+1}du=\frac{\pi}{2}$$
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.
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:
Substitute $y=\dfrac{1-z^2}{1+z^2}$ :
$$\begin{align*} & \int_0^1 \frac{\sqrt{1-y^2}}{1+y^2} \, dy \\ &= \int_0^1 \frac{4z^2}{(1+z^2)(1+z^4)} \, dz \\ &= \int_0^1 \left(-\frac2{1+z^2} + \frac{1}{1-\sqrt2\,z+z^2} + \frac{1}{1+\sqrt2\,z+z^2}\right) \, dz \\ &= -2\arctan 1 - \sqrt2 \arctan\left(1-\sqrt2\right) + \sqrt2 \arctan\left(1+\sqrt2\right) \\ &= -\frac\pi2+\frac{\sqrt2\,\pi}8 + \frac{3\sqrt2\,\pi}8 = \boxed{\frac{\sqrt2-1}2\pi} \end{align*}$$
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.
I'm not sure that this is "simpler" than trig substitution, but here's a method that makes no mention of trigonometric functions, uses a technically easier contour integration, and exploits some perhaps nonobvious symmetry of the integrand. Euler's Second Substitution, $$\sqrt{1 - y^2} = - 1 + y u,$$ rationalizes the integral: $$I := 2 \int_1^\infty \frac{(u^2 - 1)^2 \,du}{(u^2 + 1)(u^4 + 6 u^2 + 1)} .$$ Denote the integrand in $u$ by $f(u)$ for later use. The substitution $u \rightsquigarrow \frac{1}{u}$ gives that $$I = 2 \int_0^1 \frac{(u^2 - 1)^2 \,du}{(u^2 + 1)(u^4 + 6 u^2 + 1)},$$ hence $$I = \frac{1}{2} \cdot 2 \int_0^\infty \frac{(u^2 - 1)^2 \,du}{(u^2 + 1)(u^4 + 6 u^2 + 1)} .$$ As the integrand is even we have $$I = \frac{1}{2} \int_{-\infty}^\infty \frac{(u^2 - 1)^2 \,du}{(u^2 + 1)(u^4 + 6 u^2 + 1)},$$ and this latter integral is a good candidate for application of the Residue Theorem.
Let $\Gamma_R$ denote boundary of the half-disk of radius $R$ centered at the origin in the upper half-plane. For $R > 1 + \sqrt{2}$, the poles of the integrand inside $\Gamma_R$ are at $i, (\sqrt{2} \pm 1) i$, so \begin{align} \oint_{\Gamma_R} f(z) \,dz \\ &= 2 \pi i \left[\operatorname{Res}(f(z), (\sqrt{2} - 1)i) + \operatorname{Res}(f(z), i) + \operatorname{Res}(f(z), (\sqrt{2} + 1) i)\right]\\ &= 2 \pi i \left(-\frac{i}{2\sqrt{2}} + \frac{i}{2} - \frac{i}{2 \sqrt{2}} \right) \\ &= \pi (\sqrt{2} - 1) . \end{align} A standard argument shows that for large $R$ the value of the integral over the semicircular contour is $O(R^{-1})$. So, taking limits gives $$I = \frac{1}{2} \int_{-\infty}^\infty f(u) \,du = \frac{1}{2} \oint_{\Gamma_R} f(z) \,dz = \color{#bf0000}{\boxed{\frac{\pi}{2} (\sqrt{2} - 1)}} .$$
Rationalize the numerator and split it into $2$ integrals: $$\begin{align}\int_0^1\frac{\sqrt{1-x^2}}{1+x^2}\mathrm dx&=\int_0^1\frac{\mathrm dx}{(1+x^2)\sqrt{1-x^2}}-\int_0^1\frac{(1+x^2)-1}{(1+x^2)\sqrt{1-x^2}}\mathrm dx\\&=2\int_0^1\frac{\mathrm dx}{(1+x^2)\sqrt{1-x^2}}-\int_0^1\frac{\mathrm dx}{\sqrt{1-x^2}}\\&\overset{t=\frac1x}{=}2\int_1^\infty\frac{t\,\mathrm dt}{(1+t^2)\sqrt{t^2-1}}-\sin^{-1}x\Big|_0^1\\&\overset{v=\sqrt{t^2-1}}{=}2\int_0^\infty\frac{\mathrm dv}{v^2+2}-\frac\pi2\\&=\sqrt2\tan^{-1}\frac{v}{\sqrt2}\Bigg|_0^\infty-\frac\pi2\\&=\frac{\pi}{2}(\sqrt2-1)\end{align}$$
It is true that the substitutions suggested by the OP lead to another substitution and then partial fraction decomposition. But the following method, although it involves trigonometry, is much simpler than substituting $y=\sin\theta$ (or $y=\tanh u$).
By applying the following transformation formula
where $\alpha=\cos^{-1}(y)$, $y\in[0, 1]$ and $a>0$, your integral becomes
$$\begin{aligned}\int \frac{\sqrt{1-y^2}}{1+y^2}\,dy &= \int \frac{{(e^{i \alpha} - e^{-i \alpha})^2}}{{(e^{i \alpha} - e^{-i \alpha})^2 + 8}}\, d\alpha \\&= \int \frac{\sin^2(\alpha)}{\sin^2(\alpha) - 2}\, d\alpha \\&= 2\int \frac{1}{\sin^2(\alpha) - 2}\,d\alpha + \int 1\,d\alpha\\&=\int1\,d\alpha-2\int \frac{\sec^2(\alpha)}{\tan^2(\alpha)+2}\,d\alpha\qquad \left(\text{Since $\sin(\alpha)=\frac{\tan(\alpha)}{\sec(\alpha)}$.}\right)\\&=\alpha-\sqrt{2}\tan^{-1}\left(\frac{\tan(\alpha)}{\sqrt{2}}\right)+C.\qquad \left(\text{Since $\int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right) + C$.}\right) \end{aligned}$$
For the integration limits, we have that $\cos^{-1}(0)=\frac{\pi}{2}$ and $\cos^{-1}(1)=0$. Reversing the interval limits with a negative sign, we have
$$\begin{aligned}\int_{0}^{1} \frac{\sqrt{1-y^2}}{1+y^2}\,dy &= -\int_{0}^{\frac{\pi}{2}} \frac{\sin^2(\alpha)}{\sin^2(\alpha) - 2}\, d\alpha \\&= \sqrt{2} \tan^{-1}\left(\frac{\tan(\alpha)}{\sqrt{2}}\right) - \alpha\bigg|_{0}^{\frac{\pi}{2}}\\&=\lim_{{\alpha \to \frac{\pi}{2}^{-}}} \left( \sqrt{2} \tan^{-1}\left(\frac{\tan(\alpha)}{\sqrt{2}}\right) - \alpha \right)-0\\&=\color{red}{\frac{1}{2} (\sqrt{2}-1) \pi} \end{aligned}$$
Alternatively, you have the option of converting the integral
$$\int \frac{{(e^{i \alpha} - e^{-i \alpha})^2}}{{(e^{i \alpha} - e^{-i \alpha})^2 + 8}}\, d\alpha$$
into an integral of a rational function and then applying partial fraction decomposition. However, this method, although more mechanical, would not be as simple as the solution I give above.
