How do you find the integral of 3*sqrt(1-y^2)?

Question

I feel really dumb, I think it is done via substitution but I totally forgot how do this stuff.

Answer 1

No, you are not dumb. this is not your ordinary substitution.

the substitution would be y=sinx, so that you have 3*sqrt(cos^2 x) and thus you have 3secx 9under restriction to values of y. and of course the integral of sec (almost always found as example in books) is natural log (secx+tanx) etc etc

Answer 2

3*sqrt(1-y^2)

let y = sin u : u = arcsin y

dy = cos u du

1- y^2 = 1- sin^ u

= cos^2 u

sqrt(1-y^2) = sqrt(cos^2 u)

= cos u

substituting these values in the integral

[ stands fot integral

3*sqrt(1-y^2) = 3[cosu cosu du

= 3[cos^2 u du

= 3[(1 + cos 2u)/2 du

=3/2[ du + 3/2 [ cos 2u du

= 3/2 u + 3/2(sin 2u/2) + c

= 3/2(u) + 3/4 sin 2u + c

= 3/2(u) + 3/4 (2sin u cos u) + c

(3/2)u + 3/2(sin u sqrt(1- sin^2u) + c

substituting u = sin^-1 y and sin u = y

3/2(arcsin y ) + 3/2( y sqrt( 1-y^2) + c

3/2(arcsin y) + 3/2(y sqrt(1-y^2) + c

Answer 3

Actually..... You need to do it by parts. I got
(3/2)(y√(1-y²) + arcsin(y)) + C
It's the √(1-y²) term that gives it away as an inverse trig function.

Doug