What is the integral from 3 to 5 of sqrt(x^2-9) ?

The integral from 3 to 5 of sqrt(x^2-9) is -9/2 ln(3)+10

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integral from 3 to 5 of sqrt(x^2-9)

\int_{3}^{5} x^{2} - 9 d x

- \frac{9}{2} ln (3) + 10

Decimal

5.05624 \dots

Solution steps

\int_{3}^{5} x^{2} - 9 d x

Apply Trigonometric Substitution

: \int_{0}^{arcsec (\frac{5}{3})} 9 tan^{2} (u) sec (u) d u

= \int_{0}^{arcsec (\frac{5}{3})} 9 tan^{2} (u) sec (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{0}^{arcsec (\frac{5}{3})} tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 9 \cdot \int_{0}^{arcsec (\frac{5}{3})} (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 9 \cdot \int_{0}^{arcsec (\frac{5}{3})} - sec (u) + sec^{3} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (- \int_{0}^{arcsec (\frac{5}{3})} sec (u) d u + \int_{0}^{arcsec (\frac{5}{3})} sec^{3} (u) d u)

\int_{0}^{arcsec (\frac{5}{3})} sec (u) d u = ln (3)

\int_{0}^{arcsec (\frac{5}{3})} sec^{3} (u) d u = \frac{10}{9} + \frac{1}{2} ln (3)

= 9 (- ln (3) + \frac{10}{9} + \frac{1}{2} ln (3))

Simplify

9 (- ln (3) + \frac{10}{9} + \frac{1}{2} ln (3)) : - \frac{9}{2} ln (3) + 10

= - \frac{9}{2} ln (3) + 10

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What is the integral from 3 to 5 of sqrt(x^2-9) ?
The integral from 3 to 5 of sqrt(x^2-9) is -9/2 ln(3)+10