What is the integral of sqrt(x)sqrt(25-x) ?

The integral of sqrt(x)sqrt(25-x) is 625/16 (2arcsin(2/25 x-1)+sin(2arcsin(2/25 x-1)))+C

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integral of sqrt(x)sqrt(25-x)

\int x 25 - x d x

\frac{625}{16} (2 arcsin (\frac{2}{25} x - 1) + sin (2 arcsin (\frac{2}{25} x - 1))) + C

Solution steps

\int x 25 - x d x

Apply exponent rule:

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

x 25 - x = x (25 - x)

= \int x (25 - x) d x

Expand

x (25 - x) : 25 x - x^{2}

= \int 25 x - x^{2} d x

Complete the square

25 x - x^{2} : - (x - \frac{25}{2})^{2} + \frac{625}{4}

= \int - (x - \frac{25}{2})^{2} + \frac{625}{4} d x

Apply u-substitution

= \int \frac{- 4 u ^{2} + 625}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int - 4 u^{2} + 625 d u

Apply Trigonometric Substitution

= \frac{1}{2} \cdot \int \frac{625}{2} cos^{2} (v) d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{625}{2} \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{625}{2} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{625}{2} \cdot \frac{1}{2} \cdot \int 1 + cos (2 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

= \frac{1}{2} \cdot \frac{625}{2} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{2} \cdot \frac{625}{2} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= \frac{1}{2} \cdot \frac{625}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{25} (x - \frac{25}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{25} (x - \frac{25}{2}))))

Simplify

\frac{1}{2} \cdot \frac{625}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{25} (x - \frac{25}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{25} (x - \frac{25}{2})))) : \frac{625}{16} (2 arcsin (\frac{2}{25} x - 1) + sin (2 arcsin (\frac{2}{25} x - 1)))

= \frac{625}{16} (2 arcsin (\frac{2}{25} x - 1) + sin (2 arcsin (\frac{2}{25} x - 1)))

Add a constant to the solution

= \frac{625}{16} (2 arcsin (\frac{2}{25} x - 1) + sin (2 arcsin (\frac{2}{25} x - 1))) + C

What is the integral of sqrt(x)sqrt(25-x) ?
The integral of sqrt(x)sqrt(25-x) is 625/16 (2arcsin(2/25 x-1)+sin(2arcsin(2/25 x-1)))+C