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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(a^2-z^2)

\int a^{2} - z^{2} d z

\frac{1}{2} a^{2} (arcsin (\frac{z}{a}) + \frac{1}{2} sin (2 arcsin (\frac{z}{a}))) + C

Solution steps

\int a^{2} - z^{2} d z

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= a^{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

= a^{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

= a^{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)

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

Substitute back

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

= a^{2} \frac{1}{2} (arcsin (\frac{1}{a} z) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} z)))

Simplify

a^{2} \frac{1}{2} (arcsin (\frac{1}{a} z) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} z))) : \frac{1}{2} a^{2} (arcsin (\frac{z}{a}) + \frac{1}{2} sin (2 arcsin (\frac{z}{a})))

= \frac{1}{2} a^{2} (arcsin (\frac{z}{a}) + \frac{1}{2} sin (2 arcsin (\frac{z}{a})))

Add a constant to the solution

= \frac{1}{2} a^{2} (arcsin (\frac{z}{a}) + \frac{1}{2} sin (2 arcsin (\frac{z}{a}))) + C

\frac{\partial}{\partial x} (x e^{x} + 2)

y = (5 x + 2)^{5}

\frac{d}{d x} (\frac{3}{4 x ^{\frac{5}{2}}})

x^{3} + x

{\frac{1}{( s - 4 ) ^{6}}}

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