What is the integral from 0 to 1 of sqrt(3-x^2) ?

The integral from 0 to 1 of sqrt(3-x^2) is (3(2arcsin(1/(sqrt(3)))+sin(2arcsin(1/(sqrt(3))))))/4

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integral from 0 to 1 of sqrt(3-x^2)

\int_{0}^{1} 3 - x^{2} d x

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

Decimal Notation

1.63032 \dots

Solution steps

\int_{0}^{1} 3 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{1}{3})} 3 cos^{2} (u) d u

= \int_{0}^{a r c s i n (\frac{1}{3})} 3 cos^{2} (u) d u

Take the constant out:

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

= 3 \cdot \int_{0}^{a r c s i n (\frac{1}{3})} cos^{2} (u) d u

Rewrite using trig identities

= 3 \cdot \int_{0}^{a r c s i n (\frac{1}{3})} \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

= 3 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{1}{3})} 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

= 3 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{1}{3})} 1 d u + \int_{0}^{a r c s i n (\frac{1}{3})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{1}{3})} 1 d u = arcsin (\frac{1}{3})

\int_{0}^{a r c s i n (\frac{1}{3})} cos (2 u) d u = \frac{sin ( 2 arcsin ( \frac{1}{3} ) )}{2}

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

Simplify

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

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

\frac{d y}{d t} - 26 t^{12} e^{- y} = 0

\frac{2}{( x + 1 ) ^{3}}

f (x) = \frac{4 x}{x ^{2} + 1}, (1, 2)

4 e^{\frac{t}{2}}

\int \frac{1}{36 - x ^{2}} d x

What is the integral from 0 to 1 of sqrt(3-x^2) ?
The integral from 0 to 1 of sqrt(3-x^2) is (3(2arcsin(1/(sqrt(3)))+sin(2arcsin(1/(sqrt(3))))))/4