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

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

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integral of sqrt(2-x^2)

\int 2 - x^{2} d x

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

Solution steps

\int 2 - x^{2} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 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

= 2 \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

= 2 \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)

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

Substitute back

u = arcsin (\frac{1}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{2}{tan ( x )})

x^{3} + 5 x^{2} - 63 x - 140

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

\int cos (5 x) d (5 x) d x

f (x) = \frac{5}{x ^{4}}

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