Question

Find the integral of arccos x /sqrt (1-x^2)dx

Answer 1

let u=arccos x then
du = -1/sqrt{1-x^2}dx , then -du=1/sqrt{1-x^2}dx
integrate -u du =- u^2/2

Answer 2

\[\int\limits \arccos(x) / \sqrt{1 - x^2} dx = \int\limits \arccos(x) \times 1/\sqrt{1 - x^2} dx\]

this is integration by substitution

let u = arccos(x)

\[du/dx = 1/\sqrt{1-x^2}\]

then \[du = 1/\sqrt{1 - x^2} dx\]

the problem can be rewritten as

\[\int\limits u du = 1/2 u^2 + c\]

replace u with arccos(x)

gives \[\int\limits \arccos(x)/\sqrt{1 - x^2} dx = 1/2 (\arccos (x))^2 + c\]

Answer 3

oops forgot the negative