Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team

How do you integrate #int 1/sqrt(x^2-16x-7) # using trigonometric substitution?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
Jun 18, 2018

Answer:

#int dx/sqrt(x^2-16x-7) = ln abs ((x-8) + sqrt(x^2 -16x-7))+C#

Explanation:

Complete the square under the root:

#int dx/sqrt(x^2-16x-7) = int dx/sqrt((x-8)^2 -73)#

#int dx/sqrt(x^2-16x-7) = 1/sqrt73 int dx/sqrt(((x-8)/sqrt73)^2 -1)#

Substitute:

#u = (x-8)/sqrt73#

#du = dx/sqrt73#

so:

#int dx/sqrt(x^2-16x-7) = int (du)/sqrt(u^2-1)#

Restrict now to the interval #u in (1,+oo)# and substitute:

#u = cosht#

#du = sinht dt#

with #t in (0,+oo)#

#int dx/sqrt(x^2-16x-7) = int (sinht dt )/sqrt(cosh^2t-1)#

As #cosh^2t-sinh^2t = 1# and #sinht > 0# for #t>0#:

#sqrt(cosh^2t-1) = sinht#

#int dx/sqrt(x^2-16x-7) = int (sinht dt )/sinht = int dt = t+C#

undoing the substitution:

#int dx/sqrt(x^2-16x-7) = "arccosh"( (x-8)/sqrt73)+C#

and using the logarithmic form of the inverse hyperbolic cosine:

#int dx/sqrt(x^2-16x-7) = ln abs ((x-8)/sqrt73 + sqrt(((x-8)/sqrt73)^2-1))+C#

Multiplying the argument of the logarithm by #sqrt73# only changes the constant #C#, so:

#int dx/sqrt(x^2-16x-7) = ln abs ((x-8) + sqrt((x-8)^2-73))+C#

#int dx/sqrt(x^2-16x-7) = ln abs ((x-8) + sqrt(x^2 -16x-7))+C#

and by differentiating we can verify that the solution is valid also for #x-8 < -sqrt73#

Was this helpful? Let the contributor know!
1500