How do you evaluate the definite integral #int 2(pi)x(cos^(-1)(x))dx# from 0 to 1?

Question

2786 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-31T04:55:32

#int_0^1 2pixcos^-1(x)dx=pi^2/4#

Before beginning, we should know or determine that #d/dxcos^-1x=(-1)/sqrt(1-x^2)#.

Working first with the unbounded integral, we should apply integration by parts. Let:

#{(u=cos^-1x,=>,du=(-1)/sqrt(1-x^2)dx),(dv=xdx,=>,v=x^2/2):}#

Then:

#2piintxcos^-1(x)dx=pix^2cos^-1x+piintx^2/sqrt(1-x^2)dx#

On the remaining integral, let #x=sintheta# so #dx=costhetad theta#. Then:

#intx^2/sqrt(1-x^2)dx=intsin^2thetad theta=1/2int(1-cos2theta)d theta=1/2theta-1/4sin2theta#

Simplifying:

#=1/2theta-1/2sinthetacostheta=1/2sin^-1x-1/2xsqrt(1-x^2)#

Plugging this into our previous expression:

#2piintxcos^-1(x)dx=pix^2cos^-1x+pi/2sin^-1x-pi/2xsqrt(1-x^2)#

Now applying the bounds, the original integral equals:

#[pix^2cos^-1x+pi/2sin^-1x-pi/2xsqrt(1-x^2)]_0^1#

#=picos^-1(1)+pi/2sin^-1(1)-(pi/2sin^-1(0))#

#=pi/2(pi/2)=pi^2/4#