Question #9592b

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Answer 1 · 2017-03-03T00:08:43

To find the indefinite integral, let $x = sec t$ $x = sect$ ,

so that $x^{2} - 1 = {tan}^{2} x$ $x^2 -1 = tan^2x$ and $d x = sec t tan t d t$ $dx = sect tant dt$ .

Upon substitution we will get

$\int \frac{d x}{{(x^{2} - 1)}^{\frac{3}{2}}} = \int \frac{sec t tan t}{{tan}^{3} t} d t$ $int dx/(x^2-1)^(3/2) = int (sec t tant)/tan^3t dt$

$= \int sec t {cot}^{2} t d t$ $= int sect cot^2 t dt$

$= \int csc t cot t d t = - csc t + C$ $= int csct cot t dt = -csc t +C$

Since $x = sec t$ $x = sect$ , we have $cos t = \frac{1}{x}$ $cos t = 1/x$ , so

$sin t = \frac{\sqrt{x^{2} - 1}}{x}$ $sin t = sqrt(x^2-1)/x$ and

$- csc t + C = - \frac{1}{sin t} = - \frac{x}{\sqrt{x^{2} - 1}} + C$ $-csct +C= -1/sint =-x/sqrt(x^2-1)+C$

If the integrand should be $∣ ∣ x^{2} - 1 ∣ ∣$ $abs(x^2-1)$ , Then for $a < 1$ $a < 1$ , use

$∣ ∣ x^{2} - 1 ∣ ∣ = (1 - x^{2})$ $abs(x^2-1) = (1-x^2)$ .

In this case, substitute $x = sin t$ $x = sint$ and integrate to get

$\frac{x}{\sqrt{x^{2} - 1}} + C$ $x/sqrt(x^2-1)+C$