Math CentralQuandaries & Queries


Question from John:

Integral 1/(25-x^2)^3/2

Hi John,

When you see $25 - x^2$ you should immediately think of a sine substitution. First write

\[(25-x^2)^{3/2} = \left[25\left(1 - \left(\frac{x}{5}\right)^{1/2}\right) \right]^{3/2} = 125\left(1 - \left(\frac{x}{5}\right)^{1/2}\right)^{3/2}\]

Let $\large \frac{x}{5} = \sin \theta$ and then $\large \frac15 dx = \cos \theta d\theta.$ Rewrite the integrand in terms of $\theta$ and simplify using the fact that $1 - \sin^2 \theta = \cos^2 \theta.$ This will result in an easy integral which yield an answer in terms of a trig function of $\theta.$

To express the answer in terms of $x$ you can use the fact that $\large \frac{x}{5} = \sin \theta$ and hence $\cos^2 \theta = 1 - \large \left( \frac{x}{5} \right)^{2}.$


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