What is the integral of 5x^2sqrt(a^2-x^2) ?

The integral of 5x^2sqrt(a^2-x^2) is 5/8 a^4(arcsin(x/a)-1/4 sin(4arcsin(x/a)))+C

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integral of 5x^2sqrt(a^2-x^2)

\int 5 x^{2} a^{2} - x^{2} d x

\frac{5}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a}))) + C

Solution steps

\int 5 x^{2} a^{2} - x^{2} d x

Take the constant out:

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

= 5 \cdot \int x^{2} a^{2} - x^{2} d x

Apply Trigonometric Substitution

: \int a^{4} sin^{2} (u) cos^{2} (u) d u

= 5 \cdot \int a^{4} 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

= 5 a^{4} \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 5 a^{4} \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

= 5 a^{4} \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

= 5 a^{4} \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)

= 5 a^{4} \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

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

= 5 a^{4} \frac{1}{8} (arcsin (\frac{1}{a} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{a} x)))

Simplify

5 a^{4} \frac{1}{8} (arcsin (\frac{1}{a} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{a} x))) : \frac{5}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a})))

= \frac{5}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a})))

Add a constant to the solution

= \frac{5}{8} a^{4} (arcsin (\frac{x}{a}) - \frac{1}{4} sin (4 arcsin (\frac{x}{a}))) + C

x \to \frac{π}{6} lim (cot (2 x))

x \frac{d y}{d x} - y = 2020 x^{2}

\int_{0}^{x^{2}} sin (t) d t

\frac{5}{x}, \frac{5}{x ^{2}} 1, 7

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

What is the integral of 5x^2sqrt(a^2-x^2) ?
The integral of 5x^2sqrt(a^2-x^2) is 5/8 a^4(arcsin(x/a)-1/4 sin(4arcsin(x/a)))+C