How do you integrate $\int \frac{d x}{{(9 x^{2} - 25)}^{\frac{3}{2}}}$ $int dx/(9x^2-25)^(3/2)$ using trig substitutions?

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How do you integrate $\int \frac{d x}{{(9 x^{2} - 25)}^{\frac{3}{2}}}$ $int dx/(9x^2-25)^(3/2)$ using trig substitutions?

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Answer 1 · 2016-11-24T10:49:18

The answer is $= - \frac{x}{25 \cdot \sqrt{9 x^{2} - 25}} + C$ $=-x/(25*sqrt(9x^2-25))+C$

Explanation:

We use ${sec}^{2} u = 1 + {tan}^{2} u$ $sec^2u=1+tan^2u$

Let's use the substitution

$x = \frac{5 sec u}{3}$ $x=(5secu)/3$ $\Rightarrow$ $=>$ , $d x = \frac{5 sec u tan u d u}{3}$ $dx=(5secutanudu)/3$

$\int \frac{d x}{{(9 x^{2} - 25)}^{\frac{3}{2}}} = \int \frac{\frac{5 sec u tan u d u}{3}}{{(9 \cdot \frac{25}{9} {sec}^{2} u - 25)}^{\frac{3}{2}}}$ $intdx/(9x^2-25)^(3/2)=int((5secutanudu)/3)/(9*25/9sec^2u-25)^(3/2)$

$= \int \frac{5 sec u tan u d u}{3 \cdot 125 ({tan}^{3} u)}$ $=int(5secutanudu)/(3*125(tan^3u))$

$= \frac{1}{75} \int \frac{sec u d u}{{tan}^{2} u}$ $=1/75int(secudu)/tan^2u$

$= \frac{1}{75} \int \frac{{cos}^{2} u d u}{cos u {sin}^{2} u}$ $=1/75int(cos^2udu)/(cosu sin^2u)$

$= \frac{1}{75} \int \frac{cos u d u}{{sin}^{2} u}$ $=1/75int(cosudu)/(sin^2u)$

Let $v = sin u$ $v=sinu$ , $\Rightarrow$ $=>$ $d v = cos u d u$ $dv=cosudu$

Therefore,

$\frac{1}{75} \int \frac{cos u d u}{{sin}^{2} u} = \frac{1}{75} \int \frac{d v}{v^{2}}$ $1/75int(cosudu)/(sin^2u)=1/75int(dv)/v^2$

$= \frac{1}{75} \int v^{- 2} d v = \frac{1}{75} \frac{v^{- 1}}{- 1} = - \frac{1}{75 v}$ $=1/75intv^(-2)dv=1/75v^(-1)/-1=-1/(75v)$

$= - \frac{1}{75 sin u}$ $=-1/(75sinu)$

$x = \frac{5 sec u}{3}$ $x=(5secu)/3$ $\Rightarrow$ $=>$ , $cos u = \frac{5}{3 x}$ $cosu=5/(3x)$

${sin}^{2} u = 1 - {cos}^{2} u = 1 - \frac{25}{9 x^{2}} = \frac{9 x^{2} - 25}{9 x^{2}}$ $sin^2u=1-cos^2u=1-25/(9x^2)=(9x^2-25)/(9x^2)$

$\frac{1}{sin u} = \frac{3 x}{\sqrt{9 x^{2} - 25}}$ $1/sinu=(3x)/(sqrt(9x^2-25))$

Therefore,

$\int \frac{d x}{{(9 x^{2} - 25)}^{\frac{3}{2}}} = (- \frac{1}{75}) \cdot \frac{3 x}{\sqrt{9 x^{2} - 25}} + C$ $intdx/(9x^2-25)^(3/2)=(-1/75)*(3x)/(sqrt(9x^2-25))+C$

$= - \frac{x}{25 \cdot \sqrt{9 x^{2} - 25}} + C$ $=-x/(25*sqrt(9x^2-25))+C$

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How do you integrate $\int \frac{d x}{{(9 x^{2} - 25)}^{\frac{3}{2}}}$ $int dx/(9x^2-25)^(3/2)$ using trig substitutions?

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