What is the integral of sqrt(16-5x^2) ?

The integral of sqrt(16-5x^2) is 4/(sqrt(5))(2arcsin((sqrt(5))/4 x)+sin(2arcsin((sqrt(5))/4 x)))+C

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integral of sqrt(16-5x^2)

\int 16 - 5 x^{2} d x

\frac{4}{5} (2 arcsin (\frac{5}{4} x) + sin (2 arcsin (\frac{5}{4} x))) + C

Solution steps

\int 16 - 5 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{16}{5} cos^{2} (u) d u

= \int \frac{16}{5} cos^{2} (u) d u

Take the constant out:

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

= \frac{16}{5} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= \frac{16}{5} \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

= \frac{16}{5} \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

= \frac{16}{5} \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)

= \frac{16}{5} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{5}{4} x)

= \frac{16}{5} \cdot \frac{1}{2} (arcsin (\frac{5}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{5}{4} x)))

Simplify

\frac{16}{5} \cdot \frac{1}{2} (arcsin (\frac{5}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{5}{4} x))) : \frac{4}{5} (2 arcsin (\frac{5}{4} x) + sin (2 arcsin (\frac{5}{4} x)))

= \frac{4}{5} (2 arcsin (\frac{5}{4} x) + sin (2 arcsin (\frac{5}{4} x)))

Add a constant to the solution

= \frac{4}{5} (2 arcsin (\frac{5}{4} x) + sin (2 arcsin (\frac{5}{4} x))) + C

(2, 5), (- 6, - 3)

\frac{d}{d x} (\frac{sin ( x )}{2 x})

\int_{- 3}^{3} 9 - x^{2} d x

\int_{- 2}^{2} (x^{2} - 4)^{2} d x

x \to 4 lim (\frac{x - 4}{x + 5})

What is the integral of sqrt(16-5x^2) ?
The integral of sqrt(16-5x^2) is 4/(sqrt(5))(2arcsin((sqrt(5))/4 x)+sin(2arcsin((sqrt(5))/4 x)))+C