What is the integral from 0 to x of sqrt(36-t^2) ?

The integral from 0 to x of sqrt(36-t^2) is 18(arcsin(x/6)+1/2 sin(2arcsin(x/6)))

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integral from 0 to x of sqrt(36-t^2)

\int_{0}^{x} 36 - t^{2} d t

18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

Solution steps

\int_{0}^{x} 36 - t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{x}{6})} 36 cos^{2} (u) d u

= \int_{0}^{a r c s i n (\frac{x}{6})} 36 cos^{2} (u) d u

Take the constant out:

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

= 36 \cdot \int_{0}^{a r c s i n (\frac{x}{6})} cos^{2} (u) d u

Rewrite using trig identities

= 36 \cdot \int_{0}^{a r c s i n (\frac{x}{6})} \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

= 36 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{x}{6})} 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

= 36 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{x}{6})} 1 d u + \int_{0}^{a r c s i n (\frac{x}{6})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{x}{6})} 1 d u = arcsin (\frac{x}{6})

\int_{0}^{a r c s i n (\frac{x}{6})} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (\frac{x}{6}))

= 36 \cdot \frac{1}{2} (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

Simplify

36 \cdot \frac{1}{2} (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6}))) : 18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

= 18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

\frac{d}{d x} ((2 x^{5} - 3)^{8})

\frac{d}{d x} (4 e^{\frac{x}{2}})

tan^{2} (3 x) + 1

y^{^{''}} + 9 y = 3 t^{2}

π \cdot x^{\frac{1}{2}} - x^{\frac{1}{3}}

What is the integral from 0 to x of sqrt(36-t^2) ?
The integral from 0 to x of sqrt(36-t^2) is 18(arcsin(x/6)+1/2 sin(2arcsin(x/6)))