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