In the process of solving an integral, I came across the following step and couldn't proceed: $$\int \sqrt{\frac{2t^2-1}{1-2t^2+t^4}}dt$$I know that I should use partial fractions but I don't know how to apply that here. Any suggestions?

  • 4
    $\begingroup$ $$1-2t^2+t^4=(1-t^2)^2\implies \sqrt{1-2t^2+t^4}=|1-t^2|$$ Put $t=-\cos\theta$ to reach at math.stackexchange.com/questions/606251/… $\endgroup$ – lab bhattacharjee Dec 15 '13 at 4:56
  • 1
    $\begingroup$ Haha I asked that question :) I was trying to solve it, which is why I got stuck here. Been going in circles :) $\endgroup$ – Artemisia Dec 15 '13 at 5:08

Let's call $$ f(t)=\sqrt{\frac{2t^2-1}{1-2t^2+t^4}}=\frac{\sqrt{2t^2-1}}{|1-t^2|} $$ which is even because $f(-t)=f(t)$. Let's call $\varphi(t)$ the function defined for $|t|>\frac{1}{\sqrt 2}$ and $|t|\ne 1$ as $$ \varphi(t)=\frac{\sqrt{2t^2-1}}{1-t^2} $$ so that the function $f(t)$, for $|t|>\frac{1}{\sqrt 2}$, may be written as $$ f(t)=\begin{cases} \varphi(t) & |t|<1\\ -\varphi(t) & |t|>1\end{cases} $$ Then, the integral which we have to evaluate is $$ I=\int \sqrt{\frac{2t^2-1}{1-2t^2+t^4}}\operatorname{d}t=\int f(t)\operatorname{d}t=\begin{cases} \int\varphi(t) \operatorname{d}t & \frac{1}{\sqrt 2}<|t|<1\\ -\int\varphi(t)\operatorname{d}t & |t|>1\end{cases} $$ So we can evaluate the integral $$ J=\int\varphi(t) \operatorname{d}t=\int \frac{\sqrt{2t^2-1}}{1-t^2} \operatorname{d}t. $$ Let's put $$ x=\frac{\sqrt 2 t}{\sqrt{2t^2-1}} $$ so that $$\operatorname{d}t=-\frac{1}{\sqrt 2}\frac{1}{(x^2-1)^{3/2}}\operatorname{d}x$$ and $$ \begin{align} \sqrt{2t^2-1}&=\frac{1}{(x^2-1)^{1/2}} & 1-t^2&=\frac{x^2-2}{2(x^2-1)} \end{align} $$ Putting all together we find $$ J=\int \frac{1}{(x^2-1)^{1/2}}\frac{2(x^2-1)}{x^2-2}\left(-\frac{1}{\sqrt 2}\right)\frac{1}{(x^2-1)^{3/2}}\operatorname{d}x $$ and simplifying we obtain $$ J=-\sqrt 2 \int \frac{1}{(x^2-1)(x^2-2)}\operatorname{d}x. $$ Expanding the integrand function in partial fraction one has $$ \begin{align} \frac{1}{(x^2-1)(x^2-2)}&=\frac{1}{x^2-2}-\frac{1}{x^2-1}\\ &=\frac{1}{2\sqrt 2}\left(\frac{1}{x-\sqrt 2}-\frac{1}{x+\sqrt 2}\right)-\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) \end{align} $$ Integrating term by term and recalling that $\int \frac{1}{u}\operatorname{d}u=\log u + c$ we find $$ J=-\frac{1}{2}\log\left(\frac{x-\sqrt 2}{x+\sqrt 2}\right)+\frac{1}{\sqrt 2}\log\left(\frac{x-1}{x+1}\right)+C $$ Finally, substituting back $x=\frac{\sqrt 2 t}{\sqrt{2t^2-1}}$ and simplifying, we obtain $$ J=-\frac{1}{2}\log\left(\frac{t-\sqrt{2t^2-1}}{t+\sqrt{2t^2-1}}\right)+\frac{1}{\sqrt 2}\log\left(\frac{\sqrt 2 t-\sqrt{2t^2-1}}{\sqrt 2t+\sqrt{2t^2-1}}\right)+C $$

  • $\begingroup$ That is quite neat :) Thank you $\endgroup$ – Artemisia Dec 19 '13 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.