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

Answer 1

#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#

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#
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Answer 2

To integrate ( \int \frac{1}{\sqrt{x^2 - 16x + 37}} ) using trigonometric substitution, first complete the square in the denominator: ( x^2 - 16x + 37 = (x - 8)^2 + 37 - 64 = (x - 8)^2 + 9 ). Then, let ( x - 8 = 3\sec(\theta) ), which implies ( dx = 3\sec(\theta)\tan(\theta) ). Substitute these expressions into the integral, and simplify using trigonometric identities to express the integral solely in terms of ( \theta ). Finally, integrate the resulting expression with respect to ( \theta ), and then revert to the original variable ( x ) by using the inverse trigonometric functions.

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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.

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