What is the integral of sqrt(9-t^2) ?

The integral of sqrt(9-t^2) is 9/2 (arcsin(1/3 t)+1/2 sin(2arcsin(1/3 t)))+C

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integral of sqrt(9-t^2)

\int 9 - t^{2} d t

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

Solution steps

\int 9 - t^{2} d t

Apply Trigonometric Substitution

: \int 9 cos^{2} (u) d u

= \int 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 cos^{2} (u) d u

Rewrite using trig identities

= 9 \cdot \int \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 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 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 9 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{3} t)

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

Simplify

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

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

Add a constant to the solution

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

x + 2, x^{5} + 2

\frac{d}{d x} (- 2 sec^{2} (x))

x^{2} d y = (y^{2} - x y + x^{2}) d x

x + \frac{6}{x}

\frac{d}{d x} (\frac{e ^{4 x}}{x ^{20}})

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