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

Question

Cameron Alexander · Accepted Answer

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

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))#

Paisley Johnson · Answer

undefined To integrate $ \int \frac{{\text{d}x}}{{(5 - 4x - x^2)^{5/2}}} $ using trigonometric substitution:

1. Complete the square in the denominator to rewrite the expression in terms of a perfect square.
2. Apply trigonometric substitution by letting $ x = \sqrt{5}\sin(\theta) $ or $ x = \sqrt{5}\cos(\theta) $, depending on the quadratic expression under the square root.
3. Express $ \text{d}x $ in terms of $ \theta $.
4. Substitute $ x $ and $ \text{d}x $ in terms of $ \theta $ into the original integral.
5. Simplify the expression using trigonometric identities.
6. Integrate the resulting expression with respect to $ \theta $.
7. Finally, substitute back $ \theta $ in terms of $ x $ to obtain the final answer in terms of $ x $.