What is the integral of (2-8x)/(sqrt(81-4x^2)) ?

The integral of (2-8x)/(sqrt(81-4x^2)) is 2sqrt(-4x^2+81)+arcsin(2/9 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2-8x)/(sqrt(81-4x^2))

\int \frac{2 - 8 x}{81 - 4 x ^{2}} d x

2 - 4 x^{2} + 81 + arcsin (\frac{2}{9} x) + C

Solution steps

\int \frac{2 - 8 x}{81 - 4 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 18 sin (u) + 1 d u

= \int - 18 sin (u) + 1 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

= - \int 18 sin (u) d u + \int 1 d u

\int 18 sin (u) d u = - 18 cos (u)

\int 1 d u = u

= - (- 18 cos (u)) + u

Substitute back

u = arcsin (\frac{2}{9} x)

= - (- 18 cos (arcsin (\frac{2}{9} x))) + arcsin (\frac{2}{9} x)

Simplify

- (- 18 cos (arcsin (\frac{2}{9} x))) + arcsin (\frac{2}{9} x) : 2 - 4 x^{2} + 81 + arcsin (\frac{2}{9} x)

= 2 - 4 x^{2} + 81 + arcsin (\frac{2}{9} x)

Add a constant to the solution

= 2 - 4 x^{2} + 81 + arcsin (\frac{2}{9} x) + C

\frac{d x}{d t} = c_{1} x - c_{2} x^{2}

\int_{- 1}^{2} \frac{3}{x ^{2}} d x

n = 0 \sum \infty 5^{n}

(arctan (y))^{^{'}}

\int 27 x^{2} + C d x

What is the integral of (2-8x)/(sqrt(81-4x^2)) ?
The integral of (2-8x)/(sqrt(81-4x^2)) is 2sqrt(-4x^2+81)+arcsin(2/9 x)+C