Given that y = arcos(x/2) find dy/dx of arccos(x/2) and hence find the integral from 0 to 1 of arcos(x/2)dx

a)    First we can rewrite y=arccos(x/2) as cosy = x/2 which allows us to use implicit differentiation to easily differentiate each side. Taking d/dx of each side gives us -siny*(dy/dx) = 1/2 . Rearranging this expression gives dy/dx = -1/(2siny). Now we have the bottom expression in terms of siny but since y is defined as arccos(x/2) it would be useful if we could rewrite the denominator in terms of cosy. To do this we use the Pythagorean identity: (siny)^2 + (cosy)^2 =1 which can be rearranged to give siny= squareroot [1- (cosy)^2]. Now we can substitute in our expression for sin y which gives us dy/dx = -1/2* squareroot [1- (cosy)^2].   Remember that we defined cosy earlier as cosy=x/2 so now lets make that substitution which gives: dy/dx = -1/2* squareroot [1- (x/2)^2].   Finally to clean up the expression we can bring the 2 in the denominator into the square root and multiply everything out which gives: dy/dx = -1/squareroot [4-x^2]For part b of the question:    There is no obvious substitution we can make therefore we need to use integration by parts to solve it: remember the formula is: integral (u dv) = uv - integral (v du). We will first solve the integral then apply the limits to evaluate it: let us rewrite the integral of arccos(x/2) dx as the integral of [ arccos(x/2) * 1 dx ] so that we can clearly see which part we should label as u and dv. TIP: let the simpler term be = to dv because we need to integrate that to get v. So we will let u = arccos(x/2) and dv = 1 dx which gives v = x.   Remember that parts a&b of a question are often linked and require you to use your previous answer in your calculations. Here du = our answer to part a.   Applying the formula gives us: x * arccos(x/2) – (-1)integral [1/(4-x^2)] dx.     Now to evaluate the second integral we can use integration by substitution. We can let t= 4-x^2 therefore dt=-2xdx which we can rearrange to give -1/2dt = xdx.    Now rewriting the expression using t and bringing the ½ outside of the integral since it is a constant gives: x * arccos(x/2) – 1/2integral [t^-1/2] dt.   Evaluating the integral in terms of t and then replacing t with t=4-x^2 gives us x * arccos(x/2) –square root [4-x^2)].  Now to find the definite integral we need to evaluate the above expression for the limits 0 to 1; which gives arccos(1/2) – squareroot(3) – [0 – squareroot(4)] = pi/3 – squareroot (3) + squareroot (4)