What is the integral of sqrt(13-x^2) ?

The integral of sqrt(13-x^2) is 13/2 (arcsin(1/(sqrt(13))x)+1/2 sin(2arcsin(1/(sqrt(13))x)))+C

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integral of sqrt(13-x^2)

\int 13 - x^{2} d x

\frac{13}{2} (arcsin (\frac{1}{13} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{13} x))) + C

Solution steps

\int 13 - x^{2} d x

Apply Trigonometric Substitution

: \int 13 cos^{2} (u) d u

= \int 13 cos^{2} (u) d u

Take the constant out:

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

= 13 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 13 \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

= 13 \cdot \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

= 13 \cdot \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)

= 13 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{13} x)

= 13 \cdot \frac{1}{2} (arcsin (\frac{1}{13} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{13} x)))

Simplify

13 \cdot \frac{1}{2} (arcsin (\frac{1}{13} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{13} x))) : \frac{13}{2} (arcsin (\frac{1}{13} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{13} x)))

= \frac{13}{2} (arcsin (\frac{1}{13} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{13} x)))

Add a constant to the solution

= \frac{13}{2} (arcsin (\frac{1}{13} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{13} x))) + C

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

\int_{\frac{1}{5}}^{3} 14 x ln (5 x) d x

\frac{d}{d x} (d u (d, x))

\int x^{2} sec^{2} (3 x^{3}) d x

y = x^{5} ln (x)

What is the integral of sqrt(13-x^2) ?
The integral of sqrt(13-x^2) is 13/2 (arcsin(1/(sqrt(13))x)+1/2 sin(2arcsin(1/(sqrt(13))x)))+C