Have a heck of an integral here. Please Help.?

Question

((-(x-.5)^2+ (2.5*10^9)/(10^10))^.5)+.5

need to integrate this.

Ok. At least mabey an antiderivitive?
Mabey?

Answer 1

Maybe. Not mabey.

But this is the mathematics section, not the English section, so let us solve this integral:

∫((-(x-.5)^2+ (2.5*10^9)/(10^10))^.5)+.5 dx

These decimals are annoying. Let's convert them to fractions:

∫√(-(x-1/2)²+ (5/2*10^9)/(10^10))+1/2 dx

Now, let's cancel out those powers of 10:

∫√(-(x-1/2)²+ 1/4)+1/2 dx

See how much simpler it is now? Let us break this into two integrals:

∫√(1/4 - (x-1/2)²) dx + ∫1/2 dx

The integral on the right can be evaluated immediately, and for the integral on the left, let us extract a factor of 1/2 (keeping in mind that 1/2 = √(1/4)):

1/2 ∫√(1 - (2x-1)²) dx + x/2 + C

Let us now make the substitution θ = arcsin (2x-1), so that sin θ = 2x-1. Then 1/2 cos θ dθ = dx. Substituting:

1/4 ∫√(1 - sin² θ) cos θ dθ + x/2 + C

Now, 1-sin² θ = cos² θ, and for since the range of the arcsin function is -π/2 to π/2, cos θ will be nonnegative, thus √cos² θ = cos θ. Thus we have:

1/4 ∫cos² θ dθ + x/2 + C

Now, cos² θ = (cos (2θ) + 1)/2, so we have:

1/8 ∫cos (2θ)+1 dθ + x/2 + C

Integrating, we obtain:

sin (2θ)/16 + θ/8 + x/2 + C

Now, sin 2θ = 2 sin θ cos θ = 2 sin θ √(cos² θ) (since cos θ is nonnegative), and 2 sin θ √cos² θ = 2 sin θ √(1-sin² θ) = 2 (2x-1) √(1-(2x-1)²), and further θ = arcsin (2x-1), so we have:

((2x-1) √(1-(2x-1)²))/8 + arcsin (2x-1)/8 + x/2 + C

Simplifying a bit:

((2x-1) √(1-(4x² - 4x + 1)))/8 + arcsin (2x-1)/8 + x/2 + C
((2x-1) √(4x - 4x²))/8 + arcsin (2x-1)/8 + x/2 + C
((2x-1) √(x - x²))/4 + arcsin (2x-1)/8 + x/2 + C

And we are done.

Answer 2

Solve the function first then integrate.