$\int \frac{1}{(1-x^2)(2x^2-1)^{\frac{1}{4}}} dx$

Question

Let $x = \sin(t)$ which implies $dx = \cos(t)dt,$

$1 - x^2 = 1 - \sin^2(t) = \cos^2(t), 2x^2 - 1 = 2\sin^2(t) - 1 = \sin^2(t).$

So, our integral becomes:$ \int \frac{\cos(t)}{\cos^2(t)\sin^{1/4}(t)} dt $

Am I right?

Answer 1

To get rid of the most unplesant term $$\sqrt[4]{2 x^2-1}=t \quad\implies\quad x=\frac{\sqrt{t^2+1}}{\sqrt{2}}\quad\implies\quad dx=\frac{t}{\sqrt{2} \sqrt{t^2+1}}$$ which make $$I=-\sqrt 2\,\int \frac{\sqrt{t}}{\left(t^2-1\right) \sqrt{t^2+1}}\,dt$$

Now $$\frac{\sqrt{2} \sqrt{t}}{\sqrt{t^2+1}}=u\quad\implies\quad t=\frac{1+\sqrt{1-u^4}}{u^2}\quad\implies\quad dt=-\frac{2 \left(1+\sqrt{1-u^4}\right)}{u^3 \sqrt{1-u^4}}\,du$$ $$I=\int \frac{u^2}{1-u^4}\,du=\frac 12\int \frac {du}{1-u^2}-\frac 12\int \frac {du}{1+u^2}$$ Now, this is simple.