Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 20 / c / Solution

For the first definite integral, use integration by substitution $t=\tan x$, with $dx=dt/(1+t^2)$, ${\cos }^{2}x=1/(1+t^2)$ and ${\sin }^{2}x=t^2/(1+t^2)$. This givesFor the second calculation, let $D=p{\cos }^{2}x+q{\sin }^{2}x$. Since $p,q>0$, this denominator is bounded away from zero; on compact positive parameter ranges the differentiated integrands have uniform bounds. Thus differentiation under the integral sign is justified and givesAdding the two equations and using the Pythagorean trigonometric identity in the numerators yields $J=-I_p-I_q$, henceFor $p=q$, this becomes $\pi /(2p^2)$, agreeing with direct integration of the constant integrand.