# Thread: HElp on definite integrals

1. ## HElp on definite integrals

Hello,

I've done this integral in 2 steps:

1. ∫(√(2x - x²))dx = ∫(√(-((x² - 2x + 1) - 1)))dx = ∫(√(1 - (x-1)²))dx

2. Substitute u = x-1 then du/dx = 1 that means du = dx

You get:
∫(√(2x-x²))dx = ∫(√(1 - u²))du. Multiply the radicand by √(1 - u²)/√(1 - u²)

∫√(1 - u²)du = ∫((1 - u²)/√(1 - u²))du. Split the integral into a sum:

∫(1 /√(1 - u²))du - ∫((u²)/√(1 - u²))du. The first integral is arcsin(u).

The second integral has to be done by partial integration. You'll get:

∫((u²)/√(1 - u²))du = -(u/2)*√(1 - u²) + ½*arcsin(u)

Put both integrals together(be careful: between the 2 integrals is a "-" sign!):

∫(√(1 - u²))du = (u/2)*√(1 - u²) + ½*arcsin(u)

Now re-substitute:

∫(√(2x - x²))dx = ((x-1)/2)*√(1 - (x-1)²) + ½*arcsin(x-1)

EB

INT{0:2} sqrt(2x - x^2) dx

Complete the square inside the radical to get:
INT{0:2} sqrt(1 - (x - 1)^2) dx

Let sinu = x - 1 <--> cosu du = dx
INT{a:b} sqrt(1 - sin^2u)cosu du
INT{a:b} sqrt(cos^2u)cosu du
INT{a:b} cos^2u du

Let cos2u = 2cos^2u - 1 --> cos^2u = 1/2(1 + cos2u)
INT{a:b} 1/2(1 + cos2u) du
1/2 INT{a:b} (1 + cos2u) du
1/2 [u + 1/2sin2u] from {a:b}

Let sin2u = 2sinu*cosu
1/2 [u + 1/2(2sinu*cosu)] from {a:b}

Since sinu = x - 1, cosu = sqrt(1 - (x - 1)^2), u = arcsin(x - 1)
1/2 [arcsin(x - 1) + (x - 1)sqrt(1 - (x - 1)^2] from {0:2}
1/2 [(arcsin(1) + (1)sqrt(1 - 1)) - (arcsin(-1) + (-1)sqrt(1 - 1))]
1/2 [pi/2 + pi/2]
1/2 (pi) = pi/2

4. Originally Posted by ecMathGeek
INT{0:2} sqrt(2x - x^2) dx

Complete the square inside the radical to get:
INT{0:2} sqrt(1 - (x - 1)^2) dx

Let sinu = x - 1 <--> cosu du = dx
INT{a:b} sqrt(1 - sin^2u)cosu du
INT{a:b} sqrt(cos^2u)cosu du
INT{a:b} cos^2u du

Let cos2u = 2cos^2u - 1 --> cos^2u = 1/2(1 + cos2u)
INT{a:b} 1/2(1 + cos2u) du
1/2 INT{a:b} (1 + cos2u) du
1/2 [u + 1/2sin2u] from {a:b}

Let sin2u = 2sinu*cosu
1/2 [u + 1/2(2sinu*cosu)] from {a:b}

Since sinu = x - 1, cosu = sqrt(1 - (x - 1)^2), u = arcsin(x - 1)
1/2 [arcsin(x - 1) + (x - 1)sqrt(1 - (x - 1)^2] from {0:2}
1/2 [(arcsin(1) + (1)sqrt(1 - 1)) - (arcsin(-1) + (-1)sqrt(1 - 1))]
1/2 [pi/2 + pi/2]
1/2 (pi) = pi/2
oooh, that's nice. using a sin(u) substituion, i don't think i've ever seen that before. You're the man ecMathGeek!