 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/20/a/solution

For

α \neq = 0

t = αx

gives

I (α) = \int_{0}^{1} \frac{d x}{1 + α ^{2} x ^{2}} = \frac{arctan α}{α} .

α = 0

the integrand is one, so

I (0) = 1

, also the continuous limit of this expression. For nonzero

α

, differentiation under the integral sign is valid on the finite interval and yields

I^{'} (α) = - 2 α \int_{0}^{1} \frac{x ^{2}}{( 1 + α ^{2} x ^{2} ) ^{2}} d x = \frac{α / ( 1 + α ^{2} ) - a r c t a n α}{α ^{2}} .

α = 3

and use

arctan 3 = π /3

. The requested value is

\int_{0}^{1} \frac{x ^{2}}{( 1 + 3 x ^{2} ) ^{2}} d x = - \frac{I ^{'} ( 3 )}{2 3} = \frac{π}{18 3} - \frac{1}{24} = \frac{π 3}{54} - \frac{1}{24} .

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