What is the integral of 1/(pisqrt(x(1-x))) ?

The integral of 1/(pisqrt(x(1-x))) is 1/pi arcsin(2x-1)+C

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integral of 1/(pisqrt(x(1-x)))

\int \frac{1}{π x ( 1 - x )} d x

\frac{1}{π} arcsin (2 x - 1) + C

Solution steps

\int \frac{1}{π x ( 1 - x )} d x

Take the constant out:

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

= \frac{1}{π} \cdot \int \frac{1}{x ( 1 - x )} d x

Expand

x (1 - x) : x - x^{2}

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

Complete the square

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

= \frac{1}{π} \cdot \int \frac{1}{- ( x - \frac{1}{2} ) ^{2} + \frac{1}{4}} d x

Apply u-substitution

: \int \frac{2}{- 4 u ^{2} + 1} d u

= \frac{1}{π} \cdot \int \frac{2}{- 4 u ^{2} + 1} d u

Take the constant out:

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

= \frac{1}{π} \cdot 2 \cdot \int \frac{1}{- 4 u ^{2} + 1} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \frac{1}{π} \cdot 2 \cdot \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{π} \cdot 2 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{π} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{π} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 (x - \frac{1}{2}))

Simplify

\frac{1}{π} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 (x - \frac{1}{2})) : \frac{1}{π} arcsin (2 x - 1)

= \frac{1}{π} arcsin (2 x - 1)

Add a constant to the solution

= \frac{1}{π} arcsin (2 x - 1) + C

\int \frac{x ^{2}}{( x - 5 ) ^{3}} d x

\frac{d}{d x} (cos (2 - 7 x))

\frac{x}{5 - x}

f (x) = (2 x - 8)^{4} (x^{2} + x + 1)^{5}

y = x^{2} - 4 x, (5, 5)

What is the integral of 1/(pisqrt(x(1-x))) ?
The integral of 1/(pisqrt(x(1-x))) is 1/pi arcsin(2x-1)+C