What is the integral of (sqrt(1-x^2))/4 ?

The integral of (sqrt(1-x^2))/4 is 1/8 (arcsin(x)+1/2 sin(2arcsin(x)))+C

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integral of (sqrt(1-x^2))/4

\int \frac{1 - x ^{2}}{4} d x

\frac{1}{8} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) + C

Solution steps

\int \frac{1 - x ^{2}}{4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int 1 - x^{2} d x

Apply Trigonometric Substitution

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

= \frac{1}{4} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= \frac{1}{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

= \frac{1}{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

= \frac{1}{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)

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

Substitute back

u = arcsin (x)

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

Simplify

\frac{1}{4} \cdot \frac{1}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) : \frac{1}{8} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x)))

= \frac{1}{8} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x)))

Add a constant to the solution

= \frac{1}{8} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))) + C

cos^{3} (x), at x = 0

\frac{2 x ^{3}}{3} - 12 x

\frac{d}{d x} (\frac{x ^{2}}{x + 5})

f (x) = 5 x^{4}

x \to 0 lim (\frac{2 + x}{2 - x})

What is the integral of (sqrt(1-x^2))/4 ?
The integral of (sqrt(1-x^2))/4 is 1/8 (arcsin(x)+1/2 sin(2arcsin(x)))+C