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

3 Answers
Jan 2, 2018

#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#

Jan 2, 2018

#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#

Jan 2, 2018

#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 . #