What is the integral of (2-x^2)^{1/2} ?

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

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integral of (2-x^2)^{1/2}

\int (2 - x^{2})^{\frac{1}{2}} d x

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

Solution steps

\int (2 - x^{2})^{\frac{1}{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 ^{\frac{1}{2}}} x)

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = - y tan (t) + \frac{tan ( t )}{cos ( t )}

1 + 3 x, 3 \leq x \leq 9

\frac{d}{d x} (- 27 cos (3 x))

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

\int cos (\frac{n x π}{2}) d x

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