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

3 Answers
Apr 24, 2018

int \ 1/(sqrt(x^6-x^8)) \ dx = 1/4ln|(sqrt(1-x^2)-1)/(sqrt(1-x^2)+1)| - (sqrt(1-x^2))/(2x^2) + C

Explanation:

We seek:

I = int \ 1/(sqrt(x^6-x^8)) \ dx
\ \ = int \ 1/(sqrt(x^6(1-x^2))) \ dx
\ \ = int \ 1/(x^3sqrt(1-x^2)) \ dx

Consider a substitution:

u^2 = 1-x^2 => 2u(du)/dx = -2x \ \ \ , and x^2 = 1-u^2

Giving us:

I = int \ (-1/(x))/( x^3sqrt(1-x^2)) \ (-x) \ dx

\ \ = - \ int \ 1/( (x^2)^2sqrt(1-x^2)) \ (-x) \ dx

\ \ = - \ int \ 1/( (1-u^2)^2 \ u) \ u \ du

\ \ = - \ int \ 1/( (1-u^2)^2) \ du

\ \ = - \ int \ 1/( (u+1)^2(u-1)^2) \ du

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

1/( (u+1)^2(u-1)^2) -= 1/4{1/(u+1)+1/(u+1)^2-1/(u-1)+1/(u-1)^2}

Thus we can write:

-4I = int 1/(u+1)+1/(u+1)^2-1/(u-1)+1/(u-1)^2 \ du

And we can integrate to get:

-4I = ln|u+1| -1/(u+1) -ln|u-1| -1/(u-1)

:. 4I = ln|u-1| - ln|u+1| + 1/(u+1) + 1/(u-1)
\ \ \ \ \ \ \ \ = ln|(u-1)/(u+1)| + ((u-1)+(u+1))/((u+1)(u-1))
\ \ \ \ \ \ \ \ = ln|(u-1)/(u+1)| + (2u)/(u^2-1)

Then we restore the substitution:

4I = ln|(sqrt(1-x^2)-1)/(sqrt(1-x^2)+1)| + (2sqrt(1-x^2))/(1-x^2-1)

\ \ \ = ln|(sqrt(1-x^2)-1)/(sqrt(1-x^2)+1)| - (2sqrt(1-x^2))/(x^2)

Finally:

I = 1/4ln|(sqrt(1-x^2)-1)/(sqrt(1-x^2)+1)| - (sqrt(1-x^2))/(2x^2)

Apr 24, 2018

-1/2(sqrt(1-x^2)/x^2+ln((1+sqrt(1-x^2))/x))

Explanation:

color(blue)("The Trigonometric Substitution will be either " x=sinu " or " x=cosu
intdx/sqrt(x^6-x^8)=intdx/sqrt(x^6(1-x^2))

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

Applying Trigonometric Substitution

x=sinu

dx=cosu*du

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

=int(cosu*du)/(sin^3u*cosu)

Simplify

=intcsc^3udu

let I=intcsc^3udu color(green)(rarr"(1)

Apply Integration by Parts

=-intcscu*d(cotu)

=-cscucotu-(-int-cot^2ucscu*du)

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

=-cscucotu-int(csc^3u-cscu)du

=-cscucotu-intcsc^3u+intcscu

From color(green)("(1)

I=-cscucotu-I-ln(cscu+cotu)

2I=-cscucotu-ln(cscu+cotu)

I=-1/2(cscucotu+ln(cscu+cotu))

Reverse the Trigonometric Substitution

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

Simplify

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

I hope this was helpful.

Apr 24, 2018

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

Explanation:

Here,

I=int1/sqrt(x^6-x^8)dx

=int1/(x^3sqrt(1-x^2))dx

Let, x=sint=>dx=costdt

So,

I=int1/(sin^3tsqrt(1-sin^2t))costdt

=int1/(sin^3tcancelcost)cancelcostdt

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

I=intcsct*csc^2tdt

"Using "color(blue)"Integration by Parts"

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

Take, u=csct and v=csc^2t

u'=-csctcott and intvdt=-cott

:.I=csct(-cott)-int(-csctcott)(-cott)dt

=-csctcott-intcot^2tcsctdt

=-csctcott-int(csc^2t-1)csctdt

=-csctcott-intcsc^3tdt+intcsctdt

I=-csctcott-color(red)I+ln|csct-cott|+cto from color(red)((A))

:.I+I=-csctcott+ln|csct-cott|+c

2I=-1/sintxxcost/sint+ln|1/sint-cost/sint|+c

2I=-1/sintxxsqrt(1-sin^2t)/sint+ln|1/sint-sqrt(1-sin^2t)/sint|+c

Subst.back ,sint=x

2I=-1/x xxsqrt(1-x^2)/x+ln|1/x-sqrt(1-x^2)/x|+c

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

I=-sqrt(1-x^2)/(2x^2)+1/2ln|(1-sqrt(1-x^2))/x|+C, where,C=c/2