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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= x^{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

= x^{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

= x^{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)

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{4 x ^{3} - 6 x ^{4}}{2 x ^{6}} d x

y^{^{'}} = 2 y - x^{2}

(\frac{x ^{2} + x + 4}{x + 1})^{^{'}}

8 t e^{2 t}

f (x) = \frac{5 - x e ^{x}}{x + e ^{x}}

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