Question

Integration by parts for definite integrals?

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

Answer 1

The answer is `=1/2`

Explanation:

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`

Answer 2

`int_0^(1/2)arccos(2x)dx=1/2`

Explanation:

`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`

Answer 3

This solution is not by parts, but may be of interest.

Explanation:

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`

Answer 4

`int_0^(1/2)arccos(2x)dx=1/2`

Explanation:

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`