How to integrate 1/(sqrt (4 - y^2)) dy?
- 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
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
Use the fact that ∫ dx/ sqrt(a^2-x^2) = arcsin(x/a)+C
- 7 years ago
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- Anonymous1 decade ago