Question

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

Answer 1

`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)`

Answer 2

`-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.

Answer 3

`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`