Expert answers to "Evaluate integration ∫1(x2−9)32dx using trigonometry substitution"

Kyran Hudson

Answered question

2021-08-21

Evaluate integration

\int \frac{1}{{(x^{2} - 9)}^{\frac{3}{2}}} d x

using trigonometry substitution

Answer & Explanation

SkladanH

Skilled2021-08-22Added 80 answers

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

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