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How do you integrate #int sqrt(-x^2-6x-25)/xdx# using trigonometric substitution?

Answer 1

Evelyn Agard

The function: #sqrt(-x^2-6x-25)/x# is not defined in #RR#

Note that:

#-x^2-6x-25 = -(x+3)^2 -16 < 0 # for every #x#.

The function:

#sqrt(-x^2-6x-25)/x# is therefore not defined for any #x in RR#

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Answer 2

Paisley Johnson

To integrate (\int \frac{\sqrt{-x^2-6x-25}}{x} dx) using trigonometric substitution, you can let (x = -5\sin(\theta) - 3). Then, find (dx) in terms of (d\theta). After substitution, simplify the expression and integrate with respect to (\theta). Finally, convert back to the original variable (x).

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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.