What is the integral from 4 to 8 of sqrt(x^2-16) ?

The integral from 4 to 8 of sqrt(x^2-16) is 16(-1/2 ln(2+sqrt(3))+sqrt(3))

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integral from 4 to 8 of sqrt(x^2-16)

\int_{4}^{8} x^{2} - 16 d x

16 (- \frac{1}{2} ln (2 + 3) + 3)

Decimal

17.17714 \dots

Solution steps

\int_{4}^{8} x^{2} - 16 d x

Apply Trigonometric Substitution

= \int_{0}^{\frac{π}{3}} 16 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

= 16 \cdot \int_{0}^{\frac{π}{3}} tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 16 \cdot \int_{0}^{\frac{π}{3}} (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 16 \cdot \int_{0}^{\frac{π}{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

= 16 (- \int_{0}^{\frac{π}{3}} sec (u) d u + \int_{0}^{\frac{π}{3}} sec^{3} (u) d u)

\int_{0}^{\frac{π}{3}} sec (u) d u = ln (2 + 3)

\int_{0}^{\frac{π}{3}} sec^{3} (u) d u = 3 + \frac{1}{2} ln (2 + 3)

= 16 (- ln (2 + 3) + 3 + \frac{1}{2} ln (2 + 3))

Simplify

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

= 16 (- \frac{1}{2} ln (2 + 3) + 3)

What is the integral from 4 to 8 of sqrt(x^2-16) ?
The integral from 4 to 8 of sqrt(x^2-16) is 16(-1/2 ln(2+sqrt(3))+sqrt(3))