<s>[INST] Evaluating $\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\,\mathrm{d}y$; how do you avoid using a complex substitution? $\newcommand{\d}{\mathrm{d}}$The given exam question - I provide the beginning for context:

Let: $$I=\int\frac{1}{(b^2-y^2)\sqrt{c^2-y^2}}\d y$$Where $b,c\gt0$, and employ the substitution $y=\frac{cx}{\sqrt{x^2+1}}$ to show that: $$I=\int\frac{1}{b^2+(b^2-c^2)x^2}\d x$$And hence evaluate: $$\int_1^{\sqrt{2}}\frac{1}{(3-y^2)\sqrt{2-y^2}}\d y$$And: $$\int_{1/\sqrt{2}}^1\frac{y}{(3y^2-1)\sqrt{2y^2-1}}\d y$$

This I managed straightforwardly. The next part:

By means of a suitable substitution, evaluate: $$\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\d y$$

I ordinarily wouldn't have a clue about evaluating that integral, as I am not very well practised in integration (which is why I am seeking out such exercises) so I assumed they wanted us to use the same ideas. My attempt is based on the context of the question at the start, but I shouldn't have needed a complex variable substitution.

$$\begin{align}I:=&\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\d y\\\overset{y\mapsto1/y}{=}&\int_1^{\sqrt{2}}\frac{y}{(3-y^2)\sqrt{2-y^2}}\d y\end{align}$$I optimistically employ the same substitution $y=\frac{\sqrt{2}x}{\sqrt{x^2+1}}$ to get: $$\begin{align}I=&\,\sqrt{2}\int_1^\infty\frac{x}{(x^2+3)\sqrt{x^2+1}}\d x\\\overset{x\mapsto1/x}{=}&\sqrt{2}\int_0^1\frac{1}{(1+3x^2)\sqrt{1+x^2}}\d x\end{align}$$

Now, I decided that instead of trying for perhaps a very long time to come up with a similar substitution to the one at the beginning (the problem being that we have $b^2+y^2$ instead of $b^2-y^2$) I decided to enforce negative signs by introducing a complex variable substitution - which will cause some weird problem.

Let $x=\frac{iz}{\sqrt{z^2+1}}$. Then $x^2(z^2+1)=-z^2,\,z^2(x^2+1)=-x^2,\,z=\frac{-ix}{\sqrt{x^2+1}}$ (the minus sign comes from pluggin $z$ back in and seeing which of $\pm i$ works) and $\frac{\d x}{\d z}=\frac{i}{(z^2+1)\sqrt{z^2+1}}$ and we get: $$\begin{align}I&=i\sqrt{2}\int_0^{-i/\sqrt{2}}\frac{1}{1-2z^2}\d z\\&\overset{z=-\frac{iu}{\sqrt{2}}}{=}\int_0^{1}\frac{1}{1+u^2}\d u\\&=\frac{\pi}{4}\end{align}$$

However, although this isn't really complex integration, I am extremely certain we were not supposed to do it this way (since the syllabus is for pre-university mathematics) so how were we supposed to do it?
Many thanks.
 [/INST] Yes, the same kind of substitution can be used here. The substitution $t=\frac{y}{\sqrt{2y^2-1}}$ transforms the integral in question to
$$\int_1^\infty\frac{\mathrm{d}t}{t^2+1}=\tan^{-1}t\Big|_1^\infty=\frac{\pi}{4} $$
More generally, this substitution can be used to evaluate integrals of the form
$$\int\frac{\mathrm{d}x}{\left(x^2+p\right)^k\sqrt{c\left(x^2+q\right)}} $$
(See Gradshteyn & Ryzhik, Eighth edition, p. 94 for more details.)
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