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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?

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    $\begingroup$ $$1-2t^2+t^4=(1-t^2)^2\implies \sqrt{1-2t^2+t^4}=|1-t^2|$$ Put $t=-\cos\theta$ to reach at math.stackexchange.com/questions/606251/… $\endgroup$ – lab bhattacharjee Dec 15 '13 at 4:56
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    $\begingroup$ 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
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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 $$ J=-\frac{1}{2}\log\left(\frac{t-\sqrt{2t^2-1}}{t+\sqrt{2t^2-1}}\right)+\frac{1}{\sqrt 2}\log\left(\frac{\sqrt 2 t-\sqrt{2t^2-1}}{\sqrt 2t+\sqrt{2t^2-1}}\right)+C $$

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

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