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

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

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

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

\frac{1}{2} y^{2} (arcsin (\frac{x}{y}) + \frac{1}{2} sin (2 arcsin (\frac{x}{y}))) + C

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= y^{2} \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

= y^{2} \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

= y^{2} \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)

= y^{2} \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

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

= y^{2} \frac{1}{2} (arcsin (\frac{1}{y} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{y} x)))

Simplify

y^{2} \frac{1}{2} (arcsin (\frac{1}{y} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{y} x))) : \frac{1}{2} y^{2} (arcsin (\frac{x}{y}) + \frac{1}{2} sin (2 arcsin (\frac{x}{y})))

= \frac{1}{2} y^{2} (arcsin (\frac{x}{y}) + \frac{1}{2} sin (2 arcsin (\frac{x}{y})))

Add a constant to the solution

= \frac{1}{2} y^{2} (arcsin (\frac{x}{y}) + \frac{1}{2} sin (2 arcsin (\frac{x}{y}))) + C

x \to - 2 lim (25 - 5 x)

f (x) = x^{\frac{4}{3}} - 4 x^{\frac{1}{3}} + 1

\int_{0}^{π} sin (θ) d θ

\frac{d}{d x} (x^{2} 5 - x)

x^{4} sec (\frac{5}{x})

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