Evaluate the integral. Use integration by parts. Show your work. Thanks.?

Question

The integral (with upper limit 1/2 and lower limit 0) of arccos x dx.
The answer is (pi +6 - 3*sqrt of 3) / 6.

Answer 1

You need to use integration by parts to solve this problem:

int(u*dv) = u*v - int(v*du). Let:

u = arccos(x) -----------> du = -1/sqrt(1-x^2) dx
dv = dx -------------------> v= x

Now substitute in the above equation:

int(arccos[x] dx) = x*arccos[x] + int(x/sqrt(1-x^2) dx). For the 2nd integral, let:

u = 1-x^2 ----> du = -2x^dx. Thus, the integral becomes:

int([-2][-1/2]x/sqrt(1-x^2) dx) = (-1/2)int(u^[-1/2]du) =

(-1/2)(2)u^(1/2) = -u^1/2. Now substitute back in x, and you get the following integral:

int(arccos[x] dx) = x*arccos[x] - sqrt(1-x^2). Now evaluated from [0->(1/2)]

Upper limit: (1/2)*arcos(1/2) - sqrt(1-(1/2)^2) = (1/2)(pi/3) - sqrt(3/4)

= (pi/6) - sqrt(3)/2

Lower limit: (0)arccos(0) - sqrt(1-0^2) = -1

Thus, you get:

(pi/6) - sqrt(3)/2 - (-1) ------> multiply by 6 to get a common denominator, and you get:

(pi - 3sqrt(3) + 6)/6, which is your desired answer

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Hope this helps

Answer 2

OK, remember how to integrate by parts?
integral of u dv = u*v - integral(v du)

in this case, let u = arccos(x) and let dv = dx
du is then the derivative of arccos, = -1/(sqrt(1-x^2))
and v = x

Substituting, the solution is
x*arccos(x) - integral(x*(-1/sqrt(1-x^2)) dx

To solve the second (integral) term, let r = 1-x^2, then dr = -2xdx or dx = -dr/2x
so integral(x/sqrt(1-x^2))dx = integral(-1/2*1/sqrt(r))dr
=-1/2*integral(r^(-1/2) dr
= -1/2*r^.5/(.5) = -r^.5
substitute back and you get the integral = -sqrt(1-x^2) (need to evaluate at the end points)

To evaluate the first part,
x*arccos(x) = 0 at the lower limit because x = 0, and 1/2*(arccos(1/2)) = 1/2*pi/3 because cos(pi/3) is 1/2.
this is equal to pi/6.

-sqrt(1-x^2), evaluated at 1/2 is -sqrt(1-.25) = -sqrt(.75)
= -sqrt(3/4) = 1/2*sqrt(3) = 3*sqrt(3)/6.
The last conversion was simply to get a common denominator with pi/6.

-sqrt(1-x^2), evaluated at 0 is simply (-1), but we have to subtract because it is at the bottom of the integral, so it is +1 or 6/6

So, put all the terms together and you have the answer!