Integration by parts for definite integrals?

Perform the inegration by parts

#intu'v=uv-intuv'#

Reminder :

#(arccosx)'=-1/sqrt(1-u^2)#

Let #u'=1#, #=>#, #u=x#

#v=arccos(2x)#, #=>#, #v'=-2/sqrt(1-4x^2)#

Therefore,

#intarccos(2x)dx=xarccos(2x)-int((-2xdx)/sqrt(1-4x^2))#

#=xarccos(2x)-1/2sqrt(1-4x^2)+C#

So,

#int_0^(1/2)arccos(2x)dx=[xarccos(2x)-1/2sqrt(1-4x^2)]_0^(1/2)#

#=(1/2arccos(2*1/2)-1/2sqrt(1-4(1/2)^2))-(0arccos(2*0)-1/2sqrt(1-4(0)^2))#

#=1/2#

#int_0^(1/2)arccos(2x)dx#

Re write as:

#int_0^(1/2)1*arccos(2x)dx#

So #u=arccos(2x)-> u'=-2/sqrt(1-4x^2)#

#v'=1-> v=x#

Now:

#intuv'=uv-intu'v# so:

#int_0^(1/2)1*arccos(2x)dx=[xarccos(2x)]_0^(1/2)-int_0^(1/2)(-2)/sqrt(1-4x^2) x dx#

#=[xarccos(2x)]_0^(1/2)+int_0^(1/2)(2x)/sqrt(1-4x^2) dx#

To evaluate this integral:

#int_0^(1/2)(2x)/sqrt(1-4x^2) dx#

Use the substitution:

#2x=sin(u)->2dx=cos(u)du#

Also note the new limits:
#x=0-> u=0#
#x=1/2->u=pi/2#

Substituting these in, the integral now becomes:

#int_0^(pi/2)(sin(u))/sqrt(1-sin^2(u))cos(u)/2du#

Now, using:

#cos^2u+sin^2u=1 -> cos^2(u)=1-sin^2(u)#

we can simplify the expression in the square root:

#=int_0^(pi/2)(sin(u))/sqrt(cos^2u)cos(u)/2du#

#=int_0^(pi/2)(sin(u))/cos(u)cos(u)/2du=int_0^(pi/2)sin(u)/2du#

#=[-1/2cos(u)]_0^(pi/2)#

So:

#int_0^(1/2)arccos(2x)dx=[xarccos(2x)]_0^(1/2)+[-1/2cos(u)]_0^(pi/2)#

Evaluating these limits:

#[xarccos(2x)]_0^(1/2)#

#={1/2arccos(2*1/2)}-{0*arccos(1/2*0)}=0-0=0#

and

#[-1/2cos(u)]_0^(pi/2)={-1/2cos(pi/2)}-{-1/2cos(0)}#

#=0+1/2=1/2#

So the final answer will be:

#int_0^(1/2)arccos(2x)dx=[xarccos(2x)]_0^(1/2)+[-1/2cos(u)]_0^(pi/2)#

#=0+1/2=1/2#

Note that

#int_0^(1/2) arccos(2x) dx = 1/2 int_0^1 arccos(u) du#

and also note that

#y = arccos(u)# if and only if #cos(u) = y# and # 0 < u < pi#

If we turn our thinking #90^@#, we can integrate one quarter a period of the cosine.

graph{arccosx [-3.093, 3.067, -0.643, 2.438]}

The area below #y = arccos(u)# from #u = 0# to #u = 1# and above the #u# axis is exactly the same as the area left of #u = cosy# and right of the #y# axis from #y = 0# to #y = pi/2#.

That is a lot of words to explain that

#1/2 int_0^1 arccos(u) du = 1/2int_0^(pi/2) cosy dy#

# = 1/2[siny]_0^(pi/2) = 1/2[1-0] = 1/2#

Use Integration by Parts to find the indefinite integral:

#intudv=uv-intvdu#

Let:

#u=arccos(2x)=>du=-2/sqrt(1-4x^2)dx# (See proof below)

#dv=1dx=>v=x#

#--------------------#
#color(blue)(y=arccos(2x)<=>cos(y)=2x#

#color(blue)(cos(y)=2x#

Differentiate both sides implicitly (W.R.T #x#)

#color(blue)(dy/dx*-sin(y)=2#

Divide both sides by #-sin(y)#

#color(blue)(dy/dx=-2/cos(y)#

Rewrite in terms of #x#

Since #color(blue)(sin(y)=(2x)/1#

Then #color(blue)(cos(y)=sqrt(1-4x^2)/1=sqrt(1-4x^2)#

So #color(blue)(dy/dx=-2/cos(y)=-2/sqrt(1-4x^2)#

#:.color(blue)(d/dx[arccos(2x)]=-2/sqrt(1-4x^2)#

#--------------------#

Plugging in into the formula:

#(arccos(2x))(x)-int(x)(-2/sqrt(1-4x^2)dx)#

Simplify:

#xarccos(2x)+2intx/sqrt(1-4x^2)dx#

Solving for #intx/sqrt(1-4x^2)dx#

Make a substitution:

Let #u=1-4x^2=>du=-8x#

#int-1/(8sqrtu)du=>-1/8int1/sqrtudu=>-1/8intu^(-1/2)du#

Use: #color(blue)(intx^adx=(x^(a+1))/(a+1)#

#-1/8*u^(-1/2+1)/(-1/2+1)=-1/8*u^(1/2)/(1/2)=-2/8u^(1/2)=-1/4u^(1/2)#

Reverse the substitution:

#=-1/4(1-4x^2)^(1/2)=-1/4sqrt(1-4x^2)#

Going back to the problem:

#=xarccos(2x)+2[-1/4sqrt(1-4x^2)]#

#=xarccos(2x)-2/4sqrt(1-4x^2)#

#=[xarccos(2x)-1/2sqrt(1-4x^2)]_0^(1/2)#

Evaluate the upper/lower limits

#=[(color(red)(1/2))arccos(2*color(red)(1/2))-1/2sqrt(1-4(color(red)(1/2))^2)]-[(color(red)0)arccos(2*color(red)0)-1/2sqrt(1-4(color(red)0)^2)]#

#=[0]-[-1/2]=1/2#