# Which appropriate trigonometric substitution is needed to evaluate: intdx/sqrt(x^6-x^8) ? Thank you so much in advance! Greetings

Apr 24, 2018

$\int \setminus \frac{1}{\sqrt{{x}^{6} - {x}^{8}}} \setminus \mathrm{dx} = \frac{1}{4} \ln | \frac{\sqrt{1 - {x}^{2}} - 1}{\sqrt{1 - {x}^{2}} + 1} | - \frac{\sqrt{1 - {x}^{2}}}{2 {x}^{2}} + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{\sqrt{{x}^{6} - {x}^{8}}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{\sqrt{{x}^{6} \left(1 - {x}^{2}\right)}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{{x}^{3} \sqrt{1 - {x}^{2}}} \setminus \mathrm{dx}$

Consider a substitution:

${u}^{2} = 1 - {x}^{2} \implies 2 u \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x \setminus \setminus \setminus$, and ${x}^{2} = 1 - {u}^{2}$

Giving us:

$I = \int \setminus \frac{- \frac{1}{x}}{{x}^{3} \sqrt{1 - {x}^{2}}} \setminus \left(- x\right) \setminus \mathrm{dx}$

$\setminus \setminus = - \setminus \int \setminus \frac{1}{{\left({x}^{2}\right)}^{2} \sqrt{1 - {x}^{2}}} \setminus \left(- x\right) \setminus \mathrm{dx}$

$\setminus \setminus = - \setminus \int \setminus \frac{1}{{\left(1 - {u}^{2}\right)}^{2} \setminus u} \setminus u \setminus \mathrm{du}$

$\setminus \setminus = - \setminus \int \setminus \frac{1}{{\left(1 - {u}^{2}\right)}^{2}} \setminus \mathrm{du}$

$\setminus \setminus = - \setminus \int \setminus \frac{1}{{\left(u + 1\right)}^{2} {\left(u - 1\right)}^{2}} \setminus \mathrm{du}$

And we can perform a partial fraction decomposition (omitted), on the integrand, giving us:

$\frac{1}{{\left(u + 1\right)}^{2} {\left(u - 1\right)}^{2}} \equiv \frac{1}{4} \left\{\frac{1}{u + 1} + \frac{1}{u + 1} ^ 2 - \frac{1}{u - 1} + \frac{1}{u - 1} ^ 2\right\}$

Thus we can write:

$- 4 I = \int \frac{1}{u + 1} + \frac{1}{u + 1} ^ 2 - \frac{1}{u - 1} + \frac{1}{u - 1} ^ 2 \setminus \mathrm{du}$

And we can integrate to get:

$- 4 I = \ln | u + 1 | - \frac{1}{u + 1} - \ln | u - 1 | - \frac{1}{u - 1}$

$\therefore 4 I = \ln | u - 1 | - \ln | u + 1 | + \frac{1}{u + 1} + \frac{1}{u - 1}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln | \frac{u - 1}{u + 1} | + \frac{\left(u - 1\right) + \left(u + 1\right)}{\left(u + 1\right) \left(u - 1\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln | \frac{u - 1}{u + 1} | + \frac{2 u}{{u}^{2} - 1}$

Then we restore the substitution:

$4 I = \ln | \frac{\sqrt{1 - {x}^{2}} - 1}{\sqrt{1 - {x}^{2}} + 1} | + \frac{2 \sqrt{1 - {x}^{2}}}{1 - {x}^{2} - 1}$

$\setminus \setminus \setminus = \ln | \frac{\sqrt{1 - {x}^{2}} - 1}{\sqrt{1 - {x}^{2}} + 1} | - \frac{2 \sqrt{1 - {x}^{2}}}{{x}^{2}}$

Finally:

$I = \frac{1}{4} \ln | \frac{\sqrt{1 - {x}^{2}} - 1}{\sqrt{1 - {x}^{2}} + 1} | - \frac{\sqrt{1 - {x}^{2}}}{2 {x}^{2}}$

Apr 24, 2018

$- \frac{1}{2} \left(\frac{\sqrt{1 - {x}^{2}}}{x} ^ 2 + \ln \left(\frac{1 + \sqrt{1 - {x}^{2}}}{x}\right)\right)$

#### Explanation:

color(blue)("The Trigonometric Substitution will be either " x=sinu " or " x=cosu
$\int \frac{\mathrm{dx}}{\sqrt{{x}^{6} - {x}^{8}}} = \int \frac{\mathrm{dx}}{\sqrt{{x}^{6} \left(1 - {x}^{2}\right)}}$

=intdx/(x^3sqrt(1-x^2)

Applying Trigonometric Substitution

$x = \sin u$

$\mathrm{dx} = \cos u \cdot \mathrm{du}$

=intdx/(x^3sqrt(1-x^2))=int(cosu*du)/(sin^3u*sqrt(1-sin^2u)

$= \int \frac{\cos u \cdot \mathrm{du}}{{\sin}^{3} u \cdot \cos u}$

Simplify

$= \int {\csc}^{3} u \mathrm{du}$

let $I = \int {\csc}^{3} u \mathrm{du}$ color(green)(rarr"(1)

Apply Integration by Parts

$= - \int \csc u \cdot d \left(\cot u\right)$

$= - \csc u \cot u - \left(- \int - {\cot}^{2} u \csc u \cdot \mathrm{du}\right)$

color(green)(cot^2u=csc^2u-1

$= - \csc u \cot u - \int \left({\csc}^{3} u - \csc u\right) \mathrm{du}$

$= - \csc u \cot u - \int {\csc}^{3} u + \int \csc u$

From color(green)("(1)

$I = - \csc u \cot u - I - \ln \left(\csc u + \cot u\right)$

$2 I = - \csc u \cot u - \ln \left(\csc u + \cot u\right)$

$I = - \frac{1}{2} \left(\csc u \cot u + \ln \left(\csc u + \cot u\right)\right)$

Reverse the Trigonometric Substitution

I=-1/2(1/x*sqrt(1-x^2)/x+ln(1/x+sqrt(1-x^2)/x)

Simplify

$I = - \frac{1}{2} \left(\frac{\sqrt{1 - {x}^{2}}}{x} ^ 2 + \ln \left(\frac{1 + \sqrt{1 - {x}^{2}}}{x}\right)\right)$

Apr 24, 2018

$I = - \frac{\sqrt{1 - {x}^{2}}}{2 {x}^{2}} + \frac{1}{2} \ln | \frac{1 - \sqrt{1 - {x}^{2}}}{x} | + C$

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{{x}^{6} - {x}^{8}}} \mathrm{dx}$

$= \int \frac{1}{{x}^{3} \sqrt{1 - {x}^{2}}} \mathrm{dx}$

Let, $x = \sin t \implies \mathrm{dx} = \cos t \mathrm{dt}$

So,

$I = \int \frac{1}{{\sin}^{3} t \sqrt{1 - {\sin}^{2} t}} \cos t \mathrm{dt}$

$= \int \frac{1}{{\sin}^{3} t \cancel{\cos} t} \cancel{\cos} t \mathrm{dt}$

color(red)(I=intcsc^3tdt...to(A)

$I = \int \csc t \cdot {\csc}^{2} t \mathrm{dt}$

$\text{Using "color(blue)"Integration by Parts}$

color(blue)(int(u*v)dt=uintvdt-int(u'intvdt)dt

Take, $u = \csc t \mathmr{and} v = {\csc}^{2} t$

$u ' = - \csc t \cot t \mathmr{and} \int v \mathrm{dt} = - \cot t$

$\therefore I = \csc t \left(- \cot t\right) - \int \left(- \csc t \cot t\right) \left(- \cot t\right) \mathrm{dt}$

$= - \csc t \cot t - \int {\cot}^{2} t \csc t \mathrm{dt}$

$= - \csc t \cot t - \int \left({\csc}^{2} t - 1\right) \csc t \mathrm{dt}$

$= - \csc t \cot t - \int {\csc}^{3} t \mathrm{dt} + \int \csc t \mathrm{dt}$

$I = - \csc t \cot t - \textcolor{red}{I} + \ln | \csc t - \cot t | + c \to$from $\textcolor{red}{\left(A\right)}$

$\therefore I + I = - \csc t \cot t + \ln | \csc t - \cot t | + c$

$2 I = - \frac{1}{\sin} t \times \cos \frac{t}{\sin} t + \ln | \frac{1}{\sin} t - \cos \frac{t}{\sin} t | + c$

$2 I = - \frac{1}{\sin} t \times \frac{\sqrt{1 - {\sin}^{2} t}}{\sin} t + \ln | \frac{1}{\sin} t - \frac{\sqrt{1 - {\sin}^{2} t}}{\sin} t | + c$

Subst.back ,$\sin t = x$

$2 I = - \frac{1}{x} \times \frac{\sqrt{1 - {x}^{2}}}{x} + \ln | \frac{1}{x} - \frac{\sqrt{1 - {x}^{2}}}{x} | + c$

$2 I = - \frac{\sqrt{1 - {x}^{2}}}{x} ^ 2 + \ln | \frac{1 - \sqrt{1 - {x}^{2}}}{x} | + c$

$I = - \frac{\sqrt{1 - {x}^{2}}}{2 {x}^{2}} + \frac{1}{2} \ln | \frac{1 - \sqrt{1 - {x}^{2}}}{x} | + C , w h e r e , C = \frac{c}{2}$