What is the integral from 0 to x of sqrt(9-t^2) ?

The integral from 0 to x of sqrt(9-t^2) is 9/2 (arcsin(x/3)+1/2 sin(2arcsin(x/3)))

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integral from 0 to x of sqrt(9-t^2)

\int_{0}^{x} 9 - t^{2} d t

\frac{9}{2} (arcsin (\frac{x}{3}) + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

Solution steps

\int_{0}^{x} 9 - t^{2} d t

Apply Trigonometric Substitution

= \int_{0}^{a r c s i n (\frac{x}{3})} 9 cos^{2} (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}^{a r c s i n (\frac{x}{3})} cos^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int_{0}^{a r c s i n (\frac{x}{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

= 9 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{x}{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

= 9 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{x}{3})} 1 d u + \int_{0}^{a r c s i n (\frac{x}{3})} cos (2 u) d u)

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

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

= 9 \cdot \frac{1}{2} (arcsin (\frac{x}{3}) + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{x}{3}) + \frac{1}{2} sin (2 arcsin (\frac{x}{3}))) : \frac{9}{2} (arcsin (\frac{x}{3}) + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

= \frac{9}{2} (arcsin (\frac{x}{3}) + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

What is the integral from 0 to x of sqrt(9-t^2) ?
The integral from 0 to x of sqrt(9-t^2) is 9/2 (arcsin(x/3)+1/2 sin(2arcsin(x/3)))