How do you integrate #int 1/sqrt(x^2-16x+37) # using trigonometric substitution?

1 Answer
Dec 30, 2015

#ln|x-8+sqrt((x-8)^2-27)|+C#

You may also see the answer written as follows

#ln|x-8+sqrt(x^2-16x+37)|+C#

Explanation:

We begin by completing the square of the denominator.

#x^2-16x+64+37-64#

#x^2-16x+64-27#

#(x-8)^2-27#

The integral becomes

#int 1/sqrt((x-8)^2-27)dx#

Now we choose a substitution

Let #sec\theta=(x-8)/sqrt(27)#

Solve for #x#

#x-8=sqrt(27)sec\theta#

#x=sqrt(27)sec\theta+8#

Differentiate both sides

#dx=sqrt(27)sec\thetatan\thetad\theta#

Now make the substitution into the integral

#int (sqrt(27)sec\thetatan\theta)/(sqrt((sqrt(27)sec\theta+8-8)^2-27))d\theta#

#int (sqrt(27)sec\thetatan\theta)/(sqrt(27sec^2\theta-27))d\theta#

#int (sqrt(27)sec\thetatan\theta)/(sqrt(27(sec^2\theta-1)))d\theta#

#sqrt(27)/sqrt(27)int (sec\thetatan\theta)/(sqrt(tan^2\theta))d\theta#

#int (sec\thetatan\theta)/(tan\theta)d\theta#

#int sec\thetad\theta#

Now we can integrate

#ln|sec\theta+tan\theta|#

We have to get things back in terms of #x#

#sec\theta=(x-8)/sqrt(27)#

#tan\theta=(sqrt((x-8)^2-27))/sqrt(27)#

Back substituting we have

#ln|(x-8+sqrt((x-8)^2-27))/sqrt(27)|+C#

Using properties of logarithms we can rewrite

#ln|x-8+sqrt((x-8)^2-27)|-ln|sqrt(27)|+C#

#ln|sqrt(27)|# is just a constant so we can write

#ln|x-8+sqrt((x-8)^2-27)|+C#

You may also see the answer written as follows

#ln|x-8+sqrt(x^2-16x+37)|+C#