What is the integral from-6 to 6 of sqrt(64-x^2) ?

The integral from-6 to 6 of sqrt(64-x^2) is 32(2arcsin(3/4)+sin(2arcsin(3/4)))

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integral from-6 to 6 of sqrt(64-x^2)

\int_{- 6}^{6} 64 - x^{2} d x

32 (2 arcsin (\frac{3}{4}) + sin (2 arcsin (\frac{3}{4})))

Decimal

86.02498 \dots

Solution steps

\int_{- 6}^{6} 64 - x^{2} d x

Apply Trigonometric Substitution

: \int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} 64 cos^{2} (u) d u

= \int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} 64 cos^{2} (u) d u

Take the constant out:

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

= 64 \cdot \int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} cos^{2} (u) d u

Rewrite using trig identities

= 64 \cdot \int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 64 \cdot \frac{1}{2} \cdot \int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} 1 + cos (2 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

= 64 \cdot \frac{1}{2} (\int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} 1 d u + \int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} cos (2 u) d u)

\int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} 1 d u = 2 arcsin (\frac{3}{4})

\int_{- a r c s i n (\frac{3}{4})}^{a r c s i n (\frac{3}{4})} cos (2 u) d u = sin (2 arcsin (\frac{3}{4}))

= 64 \cdot \frac{1}{2} (2 arcsin (\frac{3}{4}) + sin (2 arcsin (\frac{3}{4})))

Simplify

64 \cdot \frac{1}{2} (2 arcsin (\frac{3}{4}) + sin (2 arcsin (\frac{3}{4}))) : 32 (2 arcsin (\frac{3}{4}) + sin (2 arcsin (\frac{3}{4})))

= 32 (2 arcsin (\frac{3}{4}) + sin (2 arcsin (\frac{3}{4})))

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What is the integral from-6 to 6 of sqrt(64-x^2) ?
The integral from-6 to 6 of sqrt(64-x^2) is 32(2arcsin(3/4)+sin(2arcsin(3/4)))