## Trending News

# How to integrate 1/(sqrt (4 - y^2)) dy?

### 5 Answers

- s_h_mcLv 41 decade agoFavorite Answer
Note that the derivative of arcsin(x) = 1/sqrt(1-x^2).

So, if you let y = 2u, then y^2 = 4u^2, dy = 2 du.

Our integrand becomes

2*(1/sqrt(4 - 4u^2)) du

Factor out 4 in the square root and then take the square root, you get

2*(1/(2sqrt(1-u^2))) du

The 2s in the numerator and denominator will reduce leaving

1/(sqrt(1 - u^2)) du. The integral is arcsin(u) + 1.

Back-substituting yields: arcsin(y/2) + C

- 1 decade ago
sqrt (4 - y^2) = sqrt(4(1 - (y^2)/4)) = 2sqrt(1-(y/2)^2)

let (y/2) = sin(x)

dy = cos(x) dx

2sqrt(1-(y/2)^2) = 2sqrt(1-(sin(x))^2) = 2cos(x)

int( 1/(sqrt (4 - y^2)) )dy =

int( 1/2cos(x) )cos(x) dx = int(1/2)dx = x/2 + C

x = arcsin(y/2), so int( 1/(sqrt (4 - y^2)) )dy = arcsin(y/2)/2 + C

- cidyahLv 71 decade ago
∫ dy/ sqrt(2^2 -y^2)= arcsin(y/2) +C

Note:

Use the fact that ∫ dx/ sqrt(a^2-x^2) = arcsin(x/a)+C

- How do you think about the answers? You can sign in to vote the answer.
- Anonymous1 decade ago
mathway.com