What is the integral of y^2*2(x^2-y^2)^{1/2} ?

The integral of y^2*2(x^2-y^2)^{1/2} is 1/4 x^4(arcsin(y/x)-1/4 sin(4arcsin(y/x)))+C

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integral of y^2*2(x^2-y^2)^{1/2}

integral

y^{2} \cdot 2 (x^{2} - y^{2})^{\frac{1}{2}}

\frac{1}{4} x^{4} (arcsin (\frac{y}{x}) - \frac{1}{4} sin (4 arcsin (\frac{y}{x}))) + C

Solution steps

\int y^{2} \cdot 2 (x^{2} - y^{2})^{\frac{1}{2}} d y

Take the constant out:

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

= 2 \cdot \int (x^{2} - y^{2})^{\frac{1}{2}} y^{2} d y

Apply Trigonometric Substitution

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

= 2 \cdot \int x^{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

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

Rewrite using trig identities

= 2 x^{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

= 2 x^{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

= 2 x^{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)

= 2 x^{4} \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

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

= 2 x^{4} \frac{1}{8} (arcsin (\frac{1}{x} y) - \frac{1}{4} sin (4 arcsin (\frac{1}{x} y)))

Simplify

2 x^{4} \frac{1}{8} (arcsin (\frac{1}{x} y) - \frac{1}{4} sin (4 arcsin (\frac{1}{x} y))) : \frac{1}{4} x^{4} (arcsin (\frac{y}{x}) - \frac{1}{4} sin (4 arcsin (\frac{y}{x})))

= \frac{1}{4} x^{4} (arcsin (\frac{y}{x}) - \frac{1}{4} sin (4 arcsin (\frac{y}{x})))

Add a constant to the solution

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

\frac{2 ln ( x )}{x}

\frac{1}{( y ^{2} + 6 y + 5 )}

x^{2} + 3 x

e^{x} (1 + x)

(2 x + 3)^{7}

What is the integral of y^2*2(x^2-y^2)^{1/2} ?
The integral of y^2*2(x^2-y^2)^{1/2} is 1/4 x^4(arcsin(y/x)-1/4 sin(4arcsin(y/x)))+C