Use the substitution \(\displaystyle x=\frac{\sin\theta}{a}\) to find \(\displaystyle \int_0^{1/a}\frac{x^2}{\sqrt{1-(ax)^2}}\,\mathrm{d}x\) in terms of \(a\), where \(a\) is a positive constant.[5]
Hence, or otherwise, find the exact value of \(\displaystyle \int_0^{1/2}x\cos^{-1}(2x)\,\mathrm{d}x\).[3]
Given that \(\mathrm{f}(x)=x\cos^{-1}(2x)\), find the exact area of the region enclosed by \(y=\mathrm{f}(x)\) and \(y=-\mathrm{f}(x)\).[2]
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