How do you integrate #int 1/sqrt(3x^2-12x-5)dx# using trigonometric substitution?

Answer 1

Assuming you know how to do substitution :

Complete the square, you want to have #(ax+-b)^2+-1# form
#int 1/sqrt(3x^2-12x-5)dx#
#3x^2-12x-5#
#a = sqrt(3)x# #b = 2sqrt(3)#
#a^2 = 3x^2# #b^2=12#
#3x^2-12x+12-17#
#(sqrt(3)x-2sqrt(3))^2-17#

So we have

#int 1/sqrt(3x^2-12x-5)dx = int 1/sqrt((sqrt(3)x-2sqrt(3))^2-17)dx#
#= int 1/sqrt(17(1/17(sqrt(3)x-2sqrt(3))^2-1))dx#
#= 1/sqrt17int 1/sqrt(1/17(sqrt(3)x-2sqrt(3))^2-1)dx#
In fact, at this moment you can do #u = 1/sqrt17(sqrt(3)x-2sqrt(3)) #
#du = sqrt(3/17)dx# #dx = sqrt(17/3)du#
#u^2=1/17(sqrt(3)x-2sqrt(3))^2#
#= 1/sqrt3int 1/sqrt(u^2-1)du#
You have the perfect derivate of #1/sqrt(3)arccosh(u)#
One form of the integral of #int 1/sqrt(3x^2-12x-5)dx#
is #1/sqrt(3)arccosh(1/sqrt(17)(sqrt(3)x-2sqrt(3)))#

But i don't like this form, so coming back to :

#1/sqrt17int 1/sqrt(1/17(sqrt(3)x-2sqrt(3))^2-1)dx#
#sqrt(3)x-2sqrt(3) = sqrt(17)/cos(u)#
note that #u = arccos(sqrt(17)/(sqrt(3)x-2sqrt(3)))#
#sqrt(3)dx = sqrt(17)tan(u)/cos(u)du#
#dx = sqrt(17/3)tan(u)/cos(u)#

You get

#1/sqrt3int 1/sqrt(1/(cos^2(u))-1)tan(u)/cos(u)dx#
#1/sqrt(1/(cos^2(u))-1) = 1/|tan(u)|#
#1/sqrt(3)int1/cos(u)du#
#1/sqrt(3)intcos(u)/cos^2(u)du#
let's #t = sin(u)#
#dt = cos(u)du#
#= 1/sqrt(3)int1/(1-t^2)dt#

do partial fraction to get

#1/sqrt(3)(int1/(2(t+1))dt-int1/(2(t-1))dt#)

which is

#1/(2sqrt(3))[ln(|t+1|)-ln(|t-1|)]#
Substitute back for #t = sin(u)#
#1/(2sqrt(3))[ln(|sin(u)+1|)-ln(|sin(u)-1|)]#
Substitute back for #u = arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))) #
#1/(2sqrt(3))[ln(|sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))))+1|)-ln(|sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3))))-1|)]#
assuming that #sin(arccos(sqrt(17)/(sqrt(3)x-2sqrt(3)))) = sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)#

You finally have

#1/(2sqrt(3))[ln(|sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)+1|)-ln(|sqrt(1-17/(sqrt(3) x-2 sqrt(3))^2)-1|)]#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To integrate ( \int \frac{1}{\sqrt{3x^2 - 12x - 5}} , dx ) using trigonometric substitution, follow these steps:

  1. Complete the square inside the square root: ( 3x^2 - 12x - 5 = 3(x^2 - 4x) - 5 = 3(x^2 - 4x + 4 - 4) - 5 = 3[(x - 2)^2 - 4] - 5 = 3(x - 2)^2 - 12 - 5 = 3(x - 2)^2 - 17 )

  2. Let ( x - 2 = \sqrt{17}\sec(\theta) ), so ( dx = \sqrt{17}\sec(\theta)\tan(\theta) , d\theta ).

  3. Substitute ( x - 2 = \sqrt{17}\sec(\theta) ) and ( dx = \sqrt{17}\sec(\theta)\tan(\theta) , d\theta ) into the integral.

  4. Simplify and integrate with respect to ( \theta ).

  5. Replace ( \theta ) with its original expression in terms of ( x ).

  6. Simplify the result.

This substitution allows you to express the integrand in terms of trigonometric functions, which can be integrated more easily.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7