How do you find the Integral of $(1-x^2)^(3/2)/ (x^6)$ ?

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Answer 1 · 2015-10-12T15:41:13

Go !

leeeeeeet's

$x = cos(t)$
$arccos(x) = t$

$dx = -sin(t) dt$

Integral become

$-int (sin^2(t))^(3/2)/cos^6(t)*sin(t)dt$

$-int sin^4(t)/cos^6(t)$

$-int tan^4(t)* 1/cos^2(t)dt$

Let's $u = tan(t)$

$du = 1/cos^2(t) dt$

Integral become

$-intu^4du$

$-[1/5u^5]$

Substitute back

$-[1/5tan^5(t)]$

$-[1/5tan^5(arccos(x))]$

You can simplify in

$-(1 - x^2)^(5/2)/(5 x^5)$

How do you find the Integral of (1-x^2)^(3/2)/ (x^6)(1-x^2)^(3/2)/ (x^6)?