What is the integral of sqrt(49-t^2) ?

The integral of sqrt(49-t^2) is 49/2 (arcsin(1/7 t)+1/2 sin(2arcsin(1/7 t)))+C

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integral of sqrt(49-t^2)

\int 49 - t^{2} d t

\frac{49}{2} (arcsin (\frac{1}{7} t) + \frac{1}{2} sin (2 arcsin (\frac{1}{7} t))) + C

Solution steps

\int 49 - t^{2} d t

Apply Trigonometric Substitution

: \int 49 cos^{2} (u) d u

= \int 49 cos^{2} (u) d u

Take the constant out:

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

= 49 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 49 \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

= 49 \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

= 49 \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)

= 49 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{7} t)

= 49 \cdot \frac{1}{2} (arcsin (\frac{1}{7} t) + \frac{1}{2} sin (2 arcsin (\frac{1}{7} t)))

Simplify

49 \cdot \frac{1}{2} (arcsin (\frac{1}{7} t) + \frac{1}{2} sin (2 arcsin (\frac{1}{7} t))) : \frac{49}{2} (arcsin (\frac{1}{7} t) + \frac{1}{2} sin (2 arcsin (\frac{1}{7} t)))

= \frac{49}{2} (arcsin (\frac{1}{7} t) + \frac{1}{2} sin (2 arcsin (\frac{1}{7} t)))

Add a constant to the solution

= \frac{49}{2} (arcsin (\frac{1}{7} t) + \frac{1}{2} sin (2 arcsin (\frac{1}{7} t))) + C

\frac{\partial}{\partial x} (e^{- 4 x} cos (5 x))

\frac{\partial}{\partial x} (\frac{( x - 3 y )}{x + 3 y})

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

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x})

y = x, 0, 1

What is the integral of sqrt(49-t^2) ?
The integral of sqrt(49-t^2) is 49/2 (arcsin(1/7 t)+1/2 sin(2arcsin(1/7 t)))+C