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

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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:

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

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

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

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