What is the integral of (8-x)/(sqrt(64-x^2)) ?

The integral of (8-x)/(sqrt(64-x^2)) is 8arcsin(1/8 x)+sqrt(64-x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8-x)/(sqrt(64-x^2))

\int \frac{8 - x}{64 - x ^{2}} d x

8 arcsin (\frac{1}{8} x) + 64 - x^{2} + C

Solution steps

\int \frac{8 - x}{64 - x ^{2}} d x

Expand

\frac{8 - x}{64 - x ^{2}} : \frac{8}{64 - x ^{2}} - \frac{x}{64 - x ^{2}}

= \int \frac{8}{64 - x ^{2}} - \frac{x}{64 - x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{8}{64 - x ^{2}} d x - \int \frac{x}{64 - x ^{2}} d x

\int \frac{8}{64 - x ^{2}} d x = 8 arcsin (\frac{1}{8} x)

\int \frac{x}{64 - x ^{2}} d x = - 64 - x^{2}

= 8 arcsin (\frac{1}{8} x) - (- 64 - x^{2})

Simplify

= 8 arcsin (\frac{1}{8} x) + 64 - x^{2}

Add a constant to the solution

= 8 arcsin (\frac{1}{8} x) + 64 - x^{2} + C

\int \frac{4 sin ( \frac{5 x}{3} )}{7} d x

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

\int e^{- 2 x} cos (2 x + 3) d x

\int (x^{2} + 2) (x^{3} + 6 x)^{5} d x

\int \frac{1}{49 x ^{2} + 25} d x

What is the integral of (8-x)/(sqrt(64-x^2)) ?
The integral of (8-x)/(sqrt(64-x^2)) is 8arcsin(1/8 x)+sqrt(64-x^2)+C