Paper 1, Section II, B

Complex Analysis or Complex Methods
Part IB, 2014

By choice of a suitable contour show that for a>b>0a>b>0

02πsin2θdθa+bcosθ=2πb2[aa2b2]\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]

Hence evaluate

01(1x2)1/2x2dx1+x2\int_{0}^{1} \frac{\left(1-x^{2}\right)^{1 / 2} x^{2} d x}{1+x^{2}}

using the substitution x=cos(θ/2)x=\cos (\theta / 2).