What is the integral of x^2sqrt(6^2-x^2) ?

The integral of x^2sqrt(6^2-x^2) is 162(arcsin(1/6 x)-1/4 sin(4arcsin(1/6 x)))+C

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integral of x^2sqrt(6^2-x^2)

\int x^{2} 6^{2} - x^{2} d x

162 (arcsin (\frac{1}{6} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{6} x))) + C

Solution steps

\int x^{2} 6^{2} - x^{2} d x

Apply Trigonometric Substitution

: \int 1296 sin^{2} (u) cos^{2} (u) d u

= \int 1296 sin^{2} (u) cos^{2} (u) d u

Take the constant out:

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

= 1296 \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 1296 \cdot \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= 1296 \cdot \frac{1}{8} \cdot \int 1 - cos (4 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

= 1296 \cdot \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

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

= 1296 \cdot \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

u = arcsin (\frac{1}{6} x)

= 1296 \cdot \frac{1}{8} (arcsin (\frac{1}{6} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{6} x)))

Simplify

1296 \cdot \frac{1}{8} (arcsin (\frac{1}{6} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{6} x))) : 162 (arcsin (\frac{1}{6} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{6} x)))

= 162 (arcsin (\frac{1}{6} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{6} x)))

Add a constant to the solution

= 162 (arcsin (\frac{1}{6} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{6} x))) + C

\int (x + 3) (x - 4) d x

\int_{1}^{2} x^{3} + 2 d x

\int \frac{x ^{3} - 2}{x ^{3} - x ^{2}} d x

\int \frac{1}{x ^{10}} d x

\frac{1}{( 3 - x ) ^{2}}

What is the integral of x^2sqrt(6^2-x^2) ?
The integral of x^2sqrt(6^2-x^2) is 162(arcsin(1/6 x)-1/4 sin(4arcsin(1/6 x)))+C