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

∫02πsin⁡2θdθa+bcos⁡θ=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]

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}}

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

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