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

Answer 1

#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#
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Answer 2

To evaluate the definite integral ( \int_{0}^{1} 2\pi x \cos^{-1}(x) , dx ), you can use integration by parts, where ( u = \cos^{-1}(x) ) and ( dv = 2\pi x , dx ). Then, you differentiate ( u ) to find ( du ), and integrate ( dv ) to find ( v ). Finally, apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

After evaluating the integral, substitute the limits of integration (0) and (1), and then calculate the difference to find the value of the definite integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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