What is the integral of sqrt(361-x^2) ?

The integral of sqrt(361-x^2) is 361/2 (arcsin(1/19 x)+1/2 sin(2arcsin(1/19 x)))+C

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integral of sqrt(361-x^2)

\int 361 - x^{2} d x

\frac{361}{2} (arcsin (\frac{1}{19} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{19} x))) + C

Solution steps

\int 361 - x^{2} d x

Apply Trigonometric Substitution

: \int 361 cos^{2} (u) d u

= \int 361 cos^{2} (u) d u

Take the constant out:

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

= 361 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 361 \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

= 361 \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

= 361 \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)

= 361 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{19} x)

= 361 \cdot \frac{1}{2} (arcsin (\frac{1}{19} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{19} x)))

Simplify

361 \cdot \frac{1}{2} (arcsin (\frac{1}{19} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{19} x))) : \frac{361}{2} (arcsin (\frac{1}{19} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{19} x)))

= \frac{361}{2} (arcsin (\frac{1}{19} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{19} x)))

Add a constant to the solution

= \frac{361}{2} (arcsin (\frac{1}{19} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{19} x))) + C

\int - 2 x^{- 1} d x

\int (\frac{e ^{x}}{3} + 2 x) d x

\int tan^{5} (x) \cdot sec^{3} (x) d x

\frac{d y}{d x} = - 4^{- 2} - y^{- 1} + y^{- 1} + y

y^{^{''}} - \frac{3}{x} y^{^{'}} - \frac{12}{x ^{2}} y = 4 x^{2}

What is the integral of sqrt(361-x^2) ?
The integral of sqrt(361-x^2) is 361/2 (arcsin(1/19 x)+1/2 sin(2arcsin(1/19 x)))+C