Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 12

Apply the fundamental theorem of calculus separately to the upper and lower limits. No evaluation of the integral is needed:The first result has no $y$ dependence and the second no $x$ dependence, so the mixed partial derivative is

Set $s=\sin u$, so $ds=\cos udu$, and write $J_n=I_{n,1}(\pi /2,0)$. The integration by substitution gives $J_n={\int }_0^1(1-s^3{)}^{n}ds$. For $n>0$, integration by parts yieldsBoth endpoint terms vanish. Use $s^3=1-(1-s^3)$ inside the remaining integral:ConsequentlyThis is the $p=3$ instance of the power-complement integral recurrence, with initial value $J_0=1$.

Apply the power-complement integral recurrence three times, starting from $I_{0,1}(\pi /2,0)=1$:Each factor corresponds to reducing $n$ by one; the final unreduced integral is just ${\int }_0^{\pi /2}\cos udu=1$.

Subtract the integrands first, using $\cos u-{\cos }^{3}u=\cos u{\sin }^{2}u$. Then set $s=\sin u$:The further integration by substitution $w=s^3$ makes $dw=3s^{2}ds$, soThis also holds at $n=0$.