What is the integral of sqrt(x(1-x)) ?

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

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integral of sqrt(x(1-x))

\int x (1 - x) d x

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

Solution steps

\int x (1 - x) d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (1 - x) = x (1 - x)

= \int x 1 - x d x

Apply u-substitution

: \int 2 u^{2} 1 - u^{2} d u

= \int 2 u^{2} 1 - u^{2} d u

Take the constant out:

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

= 2 \cdot \int u^{2} 1 - u^{2} d u

Apply Trigonometric Substitution

: \int sin^{2} (v) cos^{2} (v) d v

= 2 \cdot \int sin^{2} (v) cos^{2} (v) d v

Rewrite using trig identities

= 2 \cdot \int \frac{1 - cos ( 4 v )}{8} d v

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int 1 - cos (4 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{8} (\int 1 d v - \int cos (4 v) d v)

\int 1 d v = v

\int cos (4 v) d v = \frac{1}{4} sin (4 v)

= 2 \cdot \frac{1}{8} (v - \frac{1}{4} sin (4 v))

Substitute back

= 2 \cdot \frac{1}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))

Simplify

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

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

Add a constant to the solution

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

\int cos (\frac{1}{4} x) d x

- 6

n = 1 \sum \infty 2 \cdot 0. 3^{n}

\int \frac{3 x ^{2} + 4}{( x ^{2} - 2 x + 2 ) ^{2}} d x

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

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