What is the integral of sqrt(1-(x^2)/(16)) ?

The integral of sqrt(1-(x^2)/(16)) is 2arcsin(1/4 x)+sin(2arcsin(1/4 x))+C

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integral of sqrt(1-(x^2)/(16))

\int 1 - \frac{x ^{2}}{16} d x

2 arcsin (\frac{1}{4} x) + sin (2 arcsin (\frac{1}{4} x)) + C

Solution steps

\int 1 - \frac{x ^{2}}{16} d x

Apply Trigonometric Substitution

: \int 4 cos^{2} (u) d u

= \int 4 cos^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{4} x)

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

Simplify

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

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

Add a constant to the solution

= 2 arcsin (\frac{1}{4} x) + sin (2 arcsin (\frac{1}{4} x)) + C

y x \frac{d y}{d x} = \frac{sec ( 3 y )}{1 + x}

\int \frac{5 e ^{3 x}}{1 + e ^{2 x}} d x

n = 4 \sum \infty \frac{1}{( n ( n + 1 ) )}

\frac{d}{d x} (\frac{4 x}{( x ^{2} + 1 )})

\int ln (17 t) d t

What is the integral of sqrt(1-(x^2)/(16)) ?
The integral of sqrt(1-(x^2)/(16)) is 2arcsin(1/4 x)+sin(2arcsin(1/4 x))+C