What is the integral of sqrt(-x^2+4) ?

The integral of sqrt(-x^2+4) is 2arcsin(1/2 x)+sin(2arcsin(1/2 x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(-x^2+4)

\int - x^{2} + 4 d x

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

Solution steps

\int - x^{2} + 4 d x

Apply Trigonometric Substitution

: \int 4 cos^{2} (u) d u

= \int 4 cos^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

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

= 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

Simplify

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

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

Add a constant to the solution

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

\int \frac{x ^{5} + 1}{x ^{2} + 1} d x

\int 4 x + \frac{4 x}{16 - x ^{2}} d x

\int e^{4 x} sin (2 x) cos (2 x) d x

\int z z - 7 d z

\int (2 x + 1)^{6} d x

What is the integral of sqrt(-x^2+4) ?
The integral of sqrt(-x^2+4) is 2arcsin(1/2 x)+sin(2arcsin(1/2 x))+C