Integrate using trigo substitution $\int \frac{d x}{{(\sqrt{x^{2} - 4 x})}^{3}}$ $int dx/(sqrt(x^2-4x))^3$ ?

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Integrate using trigo substitution $\int \frac{d x}{{(\sqrt{x^{2} - 4 x})}^{3}}$ $int dx/(sqrt(x^2-4x))^3$ ?

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Answer 1 · 2018-07-14T18:45:17

$\frac{2 - x}{4 \sqrt{(x - 4) x} + C}$ $(2-x)/(4sqrt((x-4)x)+C$

Explanation:

At first we Substitute $u = x - 2, d u = d x$ $u=x-2,du=dx$

and we get

$\int \frac{1}{{(u^{2} - 4)}^{\frac{3}{2}}} d u$ $int1/(u^2-4)^(3/2)du$
now we substitue $u = 2 sec (s), d u = 2 tan (s) sec (s) d s$ $u=2sec(s),du=2tan(s)sec(s)ds$

then we get
$\frac{1}{4} \int cot (s) csc (s) d s = - \frac{csc (s)}{4} + C$ $1/4int cot(s)csc(s)ds=-csc(s)/4+C$
with

$s = sec (- 1) (\frac{u}{2})$ $s=sec(-1)(u/2)$ we get

$- \frac{1}{4} csc ({sec}^{- 1} (\frac{u}{2})) + C$ $-1/4csc(sec^(-1)(u/2))+C$
note that

$csc ({sec}^{- 1} (z)) = \frac{1}{\sqrt{1 - \frac{1}{z^{2}}}}$ $csc(sec^(-1)(z))=1/sqrt(1-1/z^2)$ then we get

$- \frac{u}{4 \sqrt{u^{2} - 4}} + C$ $-u/(4sqrt(u^2-4))+C$
with $u = x - 2$ $u=x-2$

we get the result
$\frac{2 - x}{4 \sqrt{(x - 4) x} + C}$ $(2-x)/(4sqrt((x-4)x)+C$

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Integrate using trigo substitution $\int \frac{d x}{{(\sqrt{x^{2} - 4 x})}^{3}}$ $int dx/(sqrt(x^2-4x))^3$ ?

1 Answer

Explanation:

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Integrate using trigo substitution int dx/(sqrt(x^2-4x))^3∫dx(√x2−4x)3int dx/(sqrt(x^2-4x))^3 ?

1 Answer

Explanation:

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Integrate using trigo substitution $\int \frac{d x}{{(\sqrt{x^{2} - 4 x})}^{3}}$ $int dx/(sqrt(x^2-4x))^3$ ?