# `int_0^(1/2) cos^(-1) x dx` Evaluate the integral

## Expert Answers `int_0^(1/2)cos^(-1)xdx`

Let's first evaluate the indefinite integral by using the method of integration by parts,

`intcos^(-1)xdx=cos^(-1)x*int1dx-int(d/dx(cos^(-1)x)int1dx)dx`

`=cos^(-1)x*x-int(-1/sqrt(1-x^2)*x)dx`

`=xcos^(-1)x+intx/sqrt(1-x^2)dx`

Now let's evaluate `intx/sqrt(1-x^2)dx` using the method of substitution,

Let's substitute `t=1-x^2`

`=>dt=-2xdx`

`intx/sqrt(1-x^2)dx=intdt/(-2sqrt(t))`

`=-1/2intdt/sqrt(t)`

`=-1/2(t^(-1/2+1)/(-1/2+1))`

`=-1/2(t^(1/2)/(1/2))`

`=-t^(1/2)`

substitute back `t=1-x^2`

`=-sqrt(1-x^2)`

`:.intcos^(-1)xdx=xcos^(-1)x-sqrt(1-x^2)+C`

Now let's evaluate the definite integral,

`int_0^(1/2)cos^(-1)x=[xcos^(-1)x-sqrt(1-x^2)]_0^(1/2)`

`=[1/2cos^(-1)1/2-sqrt(1-(1/2)^2)]-[0cos^(-1)0-sqrt(1-0^2)]`

`=[1/2*pi/3-sqrt(1-1/4)]-[-1]`

`=pi/6-sqrt(3)/2+1`

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