the integral of (sqrt(64-x^2))/x dx?

Question

the integral of (sqrt(64-x^2))/x dx

having trouble with it, tried both u sub and parts and any help in cracking this one would be greatly appreciated

Answer 1

put x=8sint
64-x^2=64(1-sin^2t)=64cos^2t
dx=8costdt
theintegral
=(8cost/8sint)8costdt
now use the half angle formula for sin and cos and integrate

Answer 2

put x=8sint, so that 64-x^2=64(1-sin^2t)=64cos^2t => sqrt(64-x^2) = 8 cos(t)
and dx=8cos(t) dt

So, yor integrand becomes (8cos(t)/8sin(t))8cos(t)dt = 8 cos^2(t)/sin(t) dt = 8 (1 - sin^2(t))/sin(t) = 8 (1/sin(t) - sin(t)) dt

We know that Int(1/sin(t) dt = -ln((cos(t) +1)/sin(t)) and Int sin(t) dt = - cos(t). Therefore your integral is -8 * ( ln((cos(t) +1)/sin(t)) + cos(t)) + C, C is a constant. Now you need some algebra to express this as a function of x. x = 8 sin(t) => t = arcsin(x/8); You substitute this in the expression in terms of t and you're done

Answer 3

try working out the square root first, you should just need to find the integral of x afterwards, which isn't so hard.

Also remember that a square root is just something to the one half power.