What is the integral of sqrt(3-2x-x^2) ?

The integral of sqrt(3-2x-x^2) is 2arcsin(1/2 (x+1))+sin(2arcsin(1/2 (x+1)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(3-2x-x^2)

\int 3 - 2 x - x^{2} d x

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

Solution steps

\int 3 - 2 x - x^{2} d x

Complete the square

3 - 2 x - x^{2} : - (x + 1)^{2} + 4

= \int - (x + 1)^{2} + 4 d x

Apply u-substitution

: \int - u^{2} + 4 d u

= \int - u^{2} + 4 d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 4 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

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 v) d v

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 v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (- 1 x^{2} - 1 y^{2} - 3 x + 2 y + 1)

\int 3 e^{7} d x

9 x^{2} + y = 9, x^{8} - y = 1

f (x) = 7 x - 2 x

\int \frac{1}{( x - 4 ) ^{5}} d x

What is the integral of sqrt(3-2x-x^2) ?
The integral of sqrt(3-2x-x^2) is 2arcsin(1/2 (x+1))+sin(2arcsin(1/2 (x+1)))+C