What is the integral of sqrt(21-x^2-4x) ?

The integral of sqrt(21-x^2-4x) is 25/2 (arcsin(1/5 (x+2))+1/2 sin(2arcsin(1/5 (x+2))))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sqrt(21-x^2-4x)

\int 21 - x^{2} - 4 x d x

\frac{25}{2} (arcsin (\frac{1}{5} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x + 2)))) + C

Solution steps

\int 21 - x^{2} - 4 x d x

Complete the square

21 - x^{2} - 4 x : - (x + 2)^{2} + 25

= \int - (x + 2)^{2} + 25 d x

Apply u-substitution

= \int - u^{2} + 25 d u

Apply Trigonometric Substitution

= \int 25 cos^{2} (v) d v

Take the constant out:

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

= 25 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 25 \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

= 25 \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

= 25 \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)

= 25 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 25 \cdot \frac{1}{2} (arcsin (\frac{1}{5} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x + 2))))

Simplify

25 \cdot \frac{1}{2} (arcsin (\frac{1}{5} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x + 2)))) : \frac{25}{2} (arcsin (\frac{1}{5} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x + 2))))

= \frac{25}{2} (arcsin (\frac{1}{5} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x + 2))))

Add a constant to the solution

= \frac{25}{2} (arcsin (\frac{1}{5} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x + 2)))) + C

What is the integral of sqrt(21-x^2-4x) ?
The integral of sqrt(21-x^2-4x) is 25/2 (arcsin(1/5 (x+2))+1/2 sin(2arcsin(1/5 (x+2))))+C