What is the integral of ((x^2))/(sqrt(25-x^2)) ?

The integral of ((x^2))/(sqrt(25-x^2)) is 25/2 (arcsin(1/5 x)-1/2 sin(2arcsin(1/5 x)))+C

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integral of ((x^2))/(sqrt(25-x^2))

\int \frac{( x ^{2} )}{25 - x ^{2}} d x

\frac{25}{2} (arcsin (\frac{1}{5} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} x))) + C

Solution steps

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

Apply Trigonometric Substitution

: \int 25 sin^{2} (u) d u

= \int 25 sin^{2} (u) d u

Take the constant out:

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

= 25 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 25 \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

= 25 \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

= 25 \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)

= 25 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{5} x)

= 25 \cdot \frac{1}{2} (arcsin (\frac{1}{5} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} x)))

Simplify

25 \cdot \frac{1}{2} (arcsin (\frac{1}{5} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} x))) : \frac{25}{2} (arcsin (\frac{1}{5} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} x)))

= \frac{25}{2} (arcsin (\frac{1}{5} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} x)))

Add a constant to the solution

= \frac{25}{2} (arcsin (\frac{1}{5} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{5} x))) + C

\int (- 9.8 x + 30) d x

csc (7 x) - csc (5 x) + sec (12 x)

y ln (x) (\frac{d y}{d x}) = (\frac{y + 1}{x})^{2}

\int_{1}^{9} x^{2} d x

\frac{x ^{2} + 3 x}{6}

What is the integral of ((x^2))/(sqrt(25-x^2)) ?
The integral of ((x^2))/(sqrt(25-x^2)) is 25/2 (arcsin(1/5 x)-1/2 sin(2arcsin(1/5 x)))+C