What is the integral of sqrt(11-x^2) ?

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

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integral of sqrt(11-x^2)

\int 11 - x^{2} d x

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

Solution steps

\int 11 - x^{2} d x

Apply Trigonometric Substitution

: \int 11 cos^{2} (u) d u

= \int 11 cos^{2} (u) d u

Take the constant out:

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

= 11 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 11 \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

= 11 \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

= 11 \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)

= 11 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (3 cos (x))

f (t) = 4 sec (9) (π t - 2)

\int cos^{4} (2 π x) d x

\int sin (n x) sin (m x) d x

f (x) = \frac{3}{2 3 x + 1}, at x = 8

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