What is the integral of sqrt(121-x^2) ?

The integral of sqrt(121-x^2) is 121/2 (arcsin(1/11 x)+1/2 sin(2arcsin(1/11 x)))+C

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integral of sqrt(121-x^2)

\int 121 - x^{2} d x

\frac{121}{2} (arcsin (\frac{1}{11} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{11} x))) + C

Solution steps

\int 121 - x^{2} d x

Apply Trigonometric Substitution

: \int 121 cos^{2} (u) d u

= \int 121 cos^{2} (u) d u

Take the constant out:

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

= 121 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 121 \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

= 121 \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

= 121 \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)

= 121 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{11} x)

= 121 \cdot \frac{1}{2} (arcsin (\frac{1}{11} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{11} x)))

Simplify

121 \cdot \frac{1}{2} (arcsin (\frac{1}{11} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{11} x))) : \frac{121}{2} (arcsin (\frac{1}{11} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{11} x)))

= \frac{121}{2} (arcsin (\frac{1}{11} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{11} x)))

Add a constant to the solution

= \frac{121}{2} (arcsin (\frac{1}{11} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{11} x))) + C

sin (6 (t - \frac{π}{6}))

x \to - \infty lim (2 x^{5})

y = \frac{1}{x}, at x = a

\frac{\partial}{\partial x} (4 tan (x y))

f (x) = tan (2 x)

What is the integral of sqrt(121-x^2) ?
The integral of sqrt(121-x^2) is 121/2 (arcsin(1/11 x)+1/2 sin(2arcsin(1/11 x)))+C