What is the integral of (3-18x)/(sqrt(64-9x^2)) ?

The integral of (3-18x)/(sqrt(64-9x^2)) is 2sqrt(-9x^2+64)+arcsin(3/8 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3-18x)/(sqrt(64-9x^2))

\int \frac{3 - 18 x}{64 - 9 x ^{2}} d x

2 - 9 x^{2} + 64 + arcsin (\frac{3}{8} x) + C

Solution steps

\int \frac{3 - 18 x}{64 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 16 sin (u) + 1 d u

= \int - 16 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 16 sin (u) d u + \int 1 d u

\int 16 sin (u) d u = - 16 cos (u)

\int 1 d u = u

= - (- 16 cos (u)) + u

Substitute back

u = arcsin (\frac{3}{8} x)

= - (- 16 cos (arcsin (\frac{3}{8} x))) + arcsin (\frac{3}{8} x)

Simplify

- (- 16 cos (arcsin (\frac{3}{8} x))) + arcsin (\frac{3}{8} x) : 2 - 9 x^{2} + 64 + arcsin (\frac{3}{8} x)

= 2 - 9 x^{2} + 64 + arcsin (\frac{3}{8} x)

Add a constant to the solution

= 2 - 9 x^{2} + 64 + arcsin (\frac{3}{8} x) + C

\frac{d y}{d x} = \frac{b ^{2} x}{a ^{2} y}

\frac{\partial}{\partial x} (e^{2 x} + y ln (x))

2 y^{^{''}} + y^{^{'}} - 10 y = 0, y (1) = 5, y^{^{'}} (1) = 2

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 5 x^{3}))

3 y^{^{'}} + 15 y = 180

What is the integral of (3-18x)/(sqrt(64-9x^2)) ?
The integral of (3-18x)/(sqrt(64-9x^2)) is 2sqrt(-9x^2+64)+arcsin(3/8 x)+C