How do you integrate #int 1/sqrt(4x^2-12x-16) # using trigonometric substitution?

Question

Scarlett Allison · Accepted Answer

# int dx/sqrt(4x^2-12x-16) = 1/2ln abs ((2x-3)+sqrt(4x^2-12x-16))+C#

Complete the square under the root at the denominator:

#int dx/sqrt(4x^2-12x-16) = int dx/sqrt( (2x-3)^2 -25)#

Substitute now:

#2x -3 = 5sect#

#dx = 5/2sect tant dt#

then:

#int dx/sqrt(4x^2-12x-16) = 5/2 int (sect tantdt)/sqrt( 25sec^2t -25)#

#int dx/sqrt(4x^2-12x-16) = 1/2 int (sect tantdt)/sqrt( sec^2t -1)#

Use now the trigonometric identity:

#sec^2t-1 = tan^2t#

Note now that the function is defined for:

#abs(2x-3) > 5#

If we restrict to the interval:

#2x-3 > 5#

then #sect# is positive and:

# sqrt( sec^2t -1) = tant#

so that:

#int dx/sqrt(4x^2-12x-16) = 1/2 int (sect tantdt)/tant = 1/2 int sectdt#

To solve this last integral there is a well known trick:

#int sect dt = int sect (sect+tant)/(sect+tant) dt#

#int sect dt = int (sec^2t+sect tant)/(sect+tant) dt#

#int sect dt = int (d(sect+ tant))/(sect+tant) dt#

#int sect dt = ln abs (sect+tant) +C#

so:

#int dx/sqrt(4x^2-12x-16) = 1/2ln abs (sect+tant) +C#

and undoing the substitution:

#int dx/sqrt(4x^2-12x-16) = 1/2ln abs ((2x-3)/5 +sqrt(((2x-3)/5)^2-1))+C#

and simplifying:

# int dx/sqrt(4x^2-12x-16) = 1/2ln abs ((2x-3)+sqrt(4x^2-12x-16))+C#

Charles Arocha · Answer

Christopher Agresta · Answer

The answer is #=1/2ln(|sqrt((2x-3)^2/25-1)+(2x-3)/5|)+C#

The denominator is

#sqrt(4x^2-12x-16)=2sqrt(x^2-3x-4)#

#=2sqrt(x^2-3x+9/4-4-9/4)#

#=2sqrt((x-3/2)^2-25/4)#

#=5sqrt(((2x-3)/5)^2-1)#

#int(dx)/sqrt(4x^2-12x-16)=1/2int(dx)/(sqrt((x-3/2)^2-25/4))#

Let #u=(2x-3)/5#, #=>#, #du=2/5dx#

Therefore,

#int(dx)/sqrt(4x^2-12x-16)=1/2int(du)/(sqrt(u^2-1))#

Let #u=sectheta#, #=>#, #du=secthetatanthetad theta#

#1/2int(du)/(sqrt(u^2-1))=1/2int(secthetatanthetad theta)/(sqrt(sec^2theta-1))#

#=1/2intsecthetad theta#

#=1/2int(sectheta(sectheta+tantheta)d theta)/(sectheta+tantheta)#

#=1/2int((sec^2theta+sethetatantheta)d theta)/(sec theta+ tan theta)#

Let #v=sectheta+tantheta#

#=>#, #dv=(sec^2theta+secthetatantheta)d theta#

And finally,

#intsecthetad theta=int(dv)/v#

#=1/2ln(v)#

#=1/2ln(sqrt(u^2-1)+u)#

#=1/2ln(|sqrt((2x-3)^2/25-1)+(2x-3)/5|)+C#

Axel Alvarez · Answer

undefined To integrate $ \int \frac{1}{\sqrt{4x^2 - 12x - 16}} $ using trigonometric substitution, you can first complete the square inside the square root. After that, you can make the substitution $ x = \frac{a}{2} \sec \theta $, where $ a $ is the coefficient of $ x $ in the square root expression. Then, you can proceed with the integration using trigonometric identities.