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?

Question

$\newcommand{\d}{\mathrm{d}}$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.

Answer 1

Here is one way to carry out the integration. Substitute $\sqrt2 y= \cosh t$

\begin{align} &\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\,{d}y\\ =&\>\frac1{\sqrt2}\int_0^{\cosh^{-1}\sqrt2} \frac1{\frac32\cosh^2t-1}dt = \frac1{\sqrt2}\int_0^{\tanh^{-1}\frac1{\sqrt2}} \frac{d(\tanh t)}{\frac12+ \tanh^2t}dt\\ =&\>\tan^{-1}\left(\sqrt2\tanh t\right)\bigg|_0^{\tanh t=\frac1{\sqrt2}}=\tan^{-1}(1)=\frac\pi4 \end{align}

Answer 2

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.)