Integral: The CAS & I

Evaluate \displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx, \alpha_2 \ne \alpha_1.



\displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx

= \displaystyle\int \frac{1}{\sqrt{-(x^2-(\alpha_1+\alpha_2)x+\alpha_1 \alpha_2)}}\;dx

= \displaystyle\int \frac{1}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1 \alpha_2}}\;dx

= \displaystyle\int \frac{1}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\left(\frac{\alpha_1+\alpha_2}{2}\right)^2 + \left(\frac{\alpha_1+\alpha_2}{2}\right)^2-\alpha_1 \alpha_2}}\;dx

= \displaystyle\int \frac{1}{\sqrt{\left(\frac{\alpha_1+\alpha_2}{2}\right)^2-\alpha_1 \alpha_2-x^2+(\alpha_1+\alpha_2)x-\left(\frac{\alpha_1+\alpha_2}{2}\right)^2}}\;dx

= \displaystyle\int \frac{1}{\sqrt{\left(\frac{\alpha_1+\alpha_2}{2}\right)^2-\alpha_1 \alpha_2-\left(x^2-(\alpha_1+\alpha_2)x+\left(\frac{\alpha_1+\alpha_2}{2}\right)^2\right)}}\;dx

= \displaystyle\int \frac{1}{\sqrt{\frac{\alpha_1^2+\alpha_2^2+2\alpha_1\alpha_2-4\alpha_1\alpha_2}{4}-\left(x^2-(\alpha_1+\alpha_2)x+\left(\frac{\alpha_1+\alpha_2}{2}\right)^2\right)}}\;dx

= \displaystyle\int \frac{1}{\sqrt{\frac{\alpha_1^2+\alpha_2^2-2\alpha_1\alpha_2}{4}-\left(x^2-(\alpha_1+\alpha_2)x+\left(\frac{\alpha_1+\alpha_2}{2}\right)^2\right)}}\;dx

= \displaystyle\int \frac{1}{\sqrt{\frac{\left(\alpha_1-\alpha_2\right)^2}{4} - \frac{4x^2-4x(\alpha_1+\alpha_2)+(\alpha_1+\alpha_2)^2}{4}}}\;dx

= \displaystyle\int \frac{2}{\sqrt{\left(\alpha_2-\alpha_1\right)^2-\left(2x-(\alpha_1+\alpha_2)\right)^2}}\;dx

For a \ne 0, \int\frac{1}{\sqrt{a^2-x^2}}\;dx = \arcsin(\frac{x}{|a|}) (see Exercise-1)

=\arcsin\left(\frac{2x-(\alpha_1+\alpha_2)}{|\alpha_2-\alpha_1|}\right)


Exercise-1 Show that for a \ne 0, \int\frac{1}{\sqrt{a^2-x^2}}\;dx = \arcsin(\frac{x}{|a|}).

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One thought on “Integral: The CAS & I

  1. Pingback: A Worthy Indefinite Integral |

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