Question

How do you integrate `int arccosx` by parts from `[0,1/2]`?

Answer 1

`I=(pi-3sqrt3)/6+1`

Explanation:

`I=int_0^(1/2)cos^(-1)xdx`

we integrate by parts

`intu(dv)/(dx)dx=uv-intv(du)/(dx)dx`

the key is to choose the appropriate values for `u " "`and `(dv)/(dx)`

`u=cos^(-1)=>(du)/(dx)=-1/sqrt(1-x^2)`

`(dv)/(dx)=1=>v=x`

`I=[xcos^(-1)x-int(-x/sqrt(1-x^2))dx]_0^(1/2)`

`I=[xcos^(-1)x+int(x/sqrt(1-x^2))dx]_0^(1/2)`

the integral can be done by inspection

`intx/sqrt(1-x^2)dx=intx(1-x^2)^(-1/2)dx`

try`d/(dx)((1-x^2)^(1/2))=1/2(-2x)(1-x^2)^(-1/2)`

`=-x(1-x^2)^(1/2)`

`:.intx(1-x^2)^(-1/2)dx=-(1-x^2)^(1/2)`

`:. I=[xcos^(-1)x-(1-x^2)^(1/2)]_0^(1/2)`

`I=[1/2cos^(-1)(1/2)-(1-(1/2)^2)^(1/2)]-[0-(1-0)^(1/2)]`

`I=1/2(pi/3)-sqrt3/2+1`

`I=(pi-3sqrt3)/6+1`

Answer 2

`int_0^(1/2)cos^-1(x)\ dx=(pi+6-3sqrt3)/6`

Explanation:

I will start by working out the antiderivative. When you have a simple function that you can't rewrite, the classic trick is to use integration by parts.

The formula for integration by parts is:
`int\ f(x)g'(x)\ dx=f(x)g(x)-int\ f'(x)g(x)\ dx`

In this case, I will let `f(x)=cos^-1(x)` and `g'(x)=1`.
`f'(x)=-1/sqrt(1-x^2)`
`g(x)=x`

Plugging this into the formula gives:
`xcos^-1(x)+int\ x/sqrt(1-x^2)\ dx`

This integral can be cracked using a u-substitution with `u=1-x^2`. The derivative of `u` is then `-2x`, so we divide through by that to integrate with respect to `u`:
`int\ cancel(x)/(-2cancel(x)sqrt(u))\ du=-int\ 1/(2sqrtu)\ du=`

This is the common derivative of `sqrtu`:
`=-sqrtu+C`

Plugging back into the original expression and resubstituting, we get:
`xcos^-1(x)-sqrt(1-x^2)+C`

Now we can plug in the limits of integration:
`int_0^(1/2)cos^-1(x)\ dx=[xcos^-1(x)-sqrt(1-x^2)]_0^(1/2)=`

`=1/2cos^-1(1/2)-sqrt(1-(1/2)^2)-(0cos^-1(0)-sqrt1)=`

`=1/2*pi/3-sqrt(1-1/4)+1=pi/6-sqrt(3/4)+1=`

`=pi/6-sqrt(3)/2+1=pi/6-(3sqrt3)/6+6/6=(pi+6-3sqrt3)/6`

Answer 3

`int_0^{1/2} arccos(x) dx = pi/6 - sqrt(3) / 2 + 1 = approx 0.657573`

Explanation:

Let's first have a look at the following integral (that we are going to need later on):

# int x / sqrt(1 - x^2) dx . #

To solve this integral, we substitute `x = cos u` and `dx = -sin(u) du`, which leads to

# [ - int cos u / sqrt(1 - cos^2 u) sin u du ]_{u = arccos x} = #

# = [ - int cos u du ]_{u = arccos x} = #

# = [ - sin u + C]_{u = arccos x} = #

# = [ - sqrt(1 - cos^2 u) + C]_{u = arccos x} = #

# = - sqrt(1 - x^2) + C . #

Now we are ready to evaluate the integral that we are actually interested in:

# int arccos(x) dx = int \underbrace{1}_{u'} * \underbrace{arccos(x)}_v dx. #

Using integration by parts, and remembering that

# d / dx arccos x = - 1 / sqrt(1 - x^2) , #

this yields

# x arccos x + int x / sqrt(1 - x^2) dx = x arccos x - sqrt(1 - x^2) + C . #

Finally, we get

# int_0^{1/2} arccos(x) dx = [ x arccos x - sqrt(1 - x^2) ]_0^{1/2} = #

# = pi/6 - sqrt(3) / 2 + 1 = approx 0.657573 . #