Question

How do you integrate `int dx/(5-4x-x^2)^(5/2)` by trigonometric substitution?

Answer 1

`int dx/(5-4x-x^2)^(5/2) = ((2+x)(19-8x-2x^2))/(243(5-4x-x^2)^(3/2))`

Explanation:

Complete the square at the denominator:

`int dx/(5-4x-x^2)^(5/2) = int dx/(9- (2+x)^2)^(5/2)`

`int dx/(5-4x-x^2)^(5/2) = 1/3^5 int dx/(1- ((2+x)/3)^2)^(5/2)`

Subtitute:

`(2+x)/3 = sint`

with `t in (-pi/2,pi/2)`

so that in the interval `cost` is positive and:

`x = -2+3sint`

`dx = 3cost dt`

Then:

`int dx/(5-4x-x^2)^(5/2) = 1/3^4 int (costdt)/(1- sin^2t)^(5/2)`

and as:

`(1-sin^2t)^(5/2) = (sqrt(1-sin^2t))^5 = cos^5t`

we have:

`int dx/(5-4x-x^2)^(5/2) = 1/3^4 int dt/cos^4t = 1/3^4 int sec^4t dt`

Solve the resulting integral using the identity `sec^2t = 1+tan^2t`:

`int sec^4t dt = int sec^2t * sec^2t dt`

`int sec^4t dt = int sec^2t (1+tan^2t) dt`

`int sec^4t dt = int sec^2t dt +int tan^2tsec^2t dt`

`int sec^4t dt = tant +tan^3t/3+C`

To undo the substitution note that:

`tant = sint/cost`

and as we noted that in the interval `cost` is positive:

`tant = sint/sqrt(1-sin^2t)`

so:

`tant = ((2+x)/3)/sqrt((1-((2+x)/3)^2)`

`tant = (2+x)/(sqrt(9-(2+x)^2)`

`tant = (2+x)/sqrt(5-4x-x^2)`

Then:

`int dx/(5-4x-x^2)^(5/2) = 1/3^4 ((2+x)/sqrt(5-4x-x^2) + (2+x)^3/(3(5-4x-x^2)^(3/2)))`

and simplifying:

`int dx/(5-4x-x^2)^(5/2) = (2+x)/(81sqrt(5-4x-x^2)) (1+ (2+x)^2/(3(5-4x-x^2)))`

`int dx/(5-4x-x^2)^(5/2) = (2+x)/(81sqrt(5-4x-x^2)) ((15-12x-3x^2 + 4+4x+x^2)/(3(5-4x-x^2)))`

`int dx/(5-4x-x^2)^(5/2) = ((2+x)(19-8x-2x^2))/(243(5-4x-x^2)^(3/2))`