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

tryd/(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 .