Paper 1, Section II, BComplex Analysis or Complex MethodsPart IB, 2014By choice of a suitable contour show that for a>b>0a>b>0a>b>0∫02πsin2θdθa+bcosθ=2πb2[a−a2−b2]\int_{0}^{2 \pi} \frac{\sin ^{2} \theta d \theta}{a+b \cos \theta}=\frac{2 \pi}{b^{2}}\left[a-\sqrt{a^{2}-b^{2}}\right]∫02πa+bcosθsin2θdθ=b22π[a−a2−b2]Hence evaluate∫01(1−x2)1/2x2dx1+x2\int_{0}^{1} \frac{\left(1-x^{2}\right)^{1 / 2} x^{2} d x}{1+x^{2}}∫011+x2(1−x2)1/2x2dxusing the substitution x=cos(θ/2)x=\cos (\theta / 2)x=cos(θ/2).