Kyran Hudson

2021-08-21

Evaluate integration
$\int \frac{1}{{\left({x}^{2}-9\right)}^{\frac{3}{2}}}dx$
using trigonometry substitution

Consider the given integral as $\int \frac{1}{{\left({x}^{2}-9\right)}^{\frac{3}{2}}}dx$
We use the trigonometric substitution $x=3\mathrm{sec}t$ to evaluate the above integral.
When $x=3\mathrm{sec}t$, the expression ${\left({x}^{2}-9\right)}^{\frac{3}{2}}$ becomes
${\left({x}^{2}-9\right)}^{\frac{3}{2}}={\left({\left(3\mathrm{sec}t\right)}^{2}-9\right)}^{\frac{3}{2}}$
$={\left(9{\mathrm{sec}}^{2}t-9\right)}^{\frac{3}{2}}$
$={\left(9\left({\mathrm{sec}}^{2}t-1\right)\right)}^{\frac{3}{2}}$
$={9}^{\frac{3}{2}}{\left({\mathrm{sec}}^{2}t-1\right)}^{\frac{3}{2}}$
$={\left({9}^{\frac{1}{2}}\right)}^{3}{\left({\mathrm{sec}}^{2}t-1\right)}^{\frac{3}{2}}$
$={3}^{3}{\left({\mathrm{tan}}^{2}t\right)}^{\frac{3}{2}}$
$=27{\left({\left({\mathrm{tan}}^{2}t\right)}^{\frac{1}{2}}\right)}^{3}$
$=27{\left(\mathrm{tan}t\right)}^{3}$
$27{\mathrm{tan}}^{3}t$
Further,
$x=3\mathrm{sec}t$
$⇒dx=d\left(\mathrm{sec}t\right)$
$⇒dx=3\mathrm{sec}t\mathrm{tan}tdt$
Substituting ${\left({x}^{2}-9\right)}^{\frac{3}{2}}=27{\mathrm{tan}}^{3}t$ and $dx=3\mathrm{sec}t\mathrm{tan}tdt$ in the integral $\int \frac{1}{{\left({x}^{2}-9\right)}^{\frac{3}{2}}}dx$ we get

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